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4. Yes. Each x-coordinate is different.
5. No. The ordered pairs (12, 7) and (12, 15) have the same x-coordinates but different y-coordiates.

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First we find solutions. Then we determine which, if any, are extraneous.


Square both sides:

Squaring the left side is as easy as it looks. Squaring an expression like the right side is where a lot of mistakes are made. It is not !! To square a two-term expression like (x-5) correctly we must use FOIL on (x-5)(x-5) or use the pattern. I prefer using the pattern:

which simplifies to:

Now that x is out of the square root we can solve for it. This is a quadratic equation so we want one side to be zero. Subtracting x and adding 3 we get:

Next we factor (or use the Quadratic Formula). This factors fairly easily:

From the Zero Product Property:
x-7 = 0 or x-4 = 0
Solving these we get:
x = 7 or x = 4

Now we check. Use the original equation:

Checking x = 7:


Check!
Checking x = 4:


Check fails!
So x = 7 is actually a solution and x = 4 is extraneous.


Square both sides:

x+4 = 3x
Subtract x:
4 = 2x
Divide by 2:
2 = x
Check.

Checking x = 2:

Check!
x = 2 is the solution to this equation. There are no extraneous solutions.

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If a number is divisible by 3 then the sum of the digits of that number are also divisible by 3. So we want the 4 and the 5 and the "four different digits" to add up to a number divisible by 3.

Probably the easiest way to do this is to pick three of the digits and then see what 4th digit would make the sum of the digits be divisible by 3. Let's say the 3 of the 4 digits are 1, 2 and 3 and let's call the one missing digit "n". So the digits are 4123n5. Now let's add the digits: 4+1+2+3+n+5. This simplifies to 15+n. So what digit (or digits) would make 15+n divisible by 3? I'll leave it up to you to figure it out. Hint: There are 4 digits that would make 15+n divisible by 3. One of them is the same as one of the digits already in the number. So there are three possible 4th digits (when you pick 1, 2 and 3 to start with) which would make the number divisible by 3 and would be different from the others.

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log^5 4 + log^5 3???

Logarithms are especially difficult to post since it is not possible to type subscripts like the bases of logarithms. Please use some English to explain what it is your logarithms are (like "base 5 log of (x+1)") or teach yourself the syntax used on Algebra.com to display logs. If you click on the "Show source" link above you will be able to see what I typed to get

to display.

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A number is divisible by 9 if the sum of the digits is divisible by 9. If we call the missing digit "n" then the sum of the digits, 8+8+n+1+2, must add up to a number which is divisible by 9.
Simplifying 8+8+n+1+2 we get 19+n. "n" must be a valid digit, 0 through 9. Which one must it be if 19+n has to be divisible by 9? There's only one possible answer. I'll leave it up to you to finish. Just try different digits for "n" until 19+n is divisible by 9.

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Since there are no xy terms and since the coefficients of the squared terms are equal, this is the equation of a circle.

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There are no unknowns/variables in what you posted. What are you/we supposed to solve for?

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You probably understand the positive exponents:
, , , , ...

To help us understand negative exponents, let's think about the opposite of the above. Instead of starting with a and repeatedly multiplying by a, think about starting with a high power of a and repeatedly dividing by a:
, , , ...
As we did this the exponents kept going down by 1. Now, let's keep going!
The exponent went down by 1 again!
again, the exponent went down by 1

etc.

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To find the LCM of some polynomials you need to see their factors. So we will factor them. First the greatest common factor:


Next, factoring patterns. The second factor of the first polynomial is a difference of squares which will factor according to the pattern . The second factor of the other polynomial fits a different pattern: or (a+b)(a+b) with the "a" being x and the "b" being 3. Using these patterns to further factor we get:



If you have trouble with LCM's I find that it can be helpful to write the factors in a special way:

2x^2-18          = 1 * 2 * (x+3) * (x-3)
5x^3+30x^2+45x   = 1     * (x+3)         * 5 * x * (x+3)
========================================================
LCM              = 1 * 2 * (x+3) * (x-3) * 5 * x * (x+3) = 10x(x+3)(x-3)(x+3)

The first lines display the polynomials and their factors. The bottom line is the LCM. Notice how I used spacing and the Commutative Property (which allows me to rearrange the order of factors of the polynomials in any way I choose. I arranged the factors so the the factors in each column were all the same. I lined up the factors of 1, of 2, of (x+3), etc. Taking 1 factor from each column, the LCM is the product of these factors.

Sometimes you want an LCM in factored form (like when you are finding a lowest common denominator (LCD) which is just an LCM of denominators). Sometimes you want it in simplified form.

If you want this LCM in simplified form then we just multiply out the factors. We can multiply in any order so I am going to choose an order that looks easy. First, (x+3)(x-3). We already know how this works out from when we factored 2x^2=18:
LCM =
Next I'll use FOIL to multiply the last two factors:
LCM =
Simplifying...
LCM =
And finally the 10x (with the Distributive Property):
LCM =

P.S. There are additional uses for arranging the factors in a table like the one above:

• This table will handle as many polynomials as you need. Just add a line for each additional polynomial.
• If you are looking for an LCD, not only can you find the LCD but you can see what you need to multiply each fraction by to make the denominator become the LCD. For example, if the problem was to add:
  
  the LCD would be as was found earlier. And from the table you can literally see which factors of the LCD a certain denominator is missing. For 2x^2-18 we can see that it is "missing" factors of 5, x and a second (x+3). And for the other denominator we can see that, compared to the LCM/LCD, it is "missing" factors of 2 and (x-3). We now now how to change the fractions so their denominators become the LCD:
  etc.
• This table of factors can also be used to find the greatest common factor (GCF) between polynomials. In this case, instead of using factors from every column you use only the factors from "full" columns. IOW from columns that have no empty spaces. In this case there are only two "full" columns: the 1's and the first (x+3)'s. So the GCF for 2x^2-18 and 5x^3+30x^2+45x is 1*(x+3) or just (x+3)

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I just solved a problem very much like this one. The steps to its solution should help you figure out this problem. Click here to see the other solution.

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First, isolate the absolute value. Subtracting 4:

and divide by 3:


With the absolute value isolated, we now rewrite the equation as two equivalent equations without absolute value. If the absolute value of something is 50 then it must be either a 50 or a -50:
or
With the x out of the absolute value, we can now solve for it. Subtracting 5 in each equation:
or
Dividing (or multiplying) by -1:
or

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If "n" is the missing frequency then:

Simplifying...


Eliminate the fraction by multiplying each side by (47+n):


Subtracting 15.38n:

Subtracting 702:

Dividing by 2.62:


Since "n" was a frequency, it should be a whole number. The reason we did not get a whole number for n must be either

• We made a mistake. Check the work to see if we did. Or...
• The average we were given is a rounded-off number. In this case, "n" must be 8.

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With all the equations with a zero on the right side, we know that a solution (perhaps the only solution) to this system will be (0, 0, 0). But we will go ahead and find an inverse anyway.

The coefficient matrix for this system is:

1   3  -1
2  -1   4
1 -11  14

(Note: To save time I am not going to put the matrices in brackets. So you'll just have to deal with just the numbers.) To find the inverse of this matrix:

1. "Append" the identity matrix to the coefficient matrix. This should give you a matrix with the left half being the original coefficient matrix and the right half being the identity matrix.
2. Use elementary row operations to change this expanded matrix so the the left half is the identity matrix.
3. The right half of this modified, expanded matrix will be the inverse.

Let's see this in action:
1. Append the identity matrix:

1   3  -1   1   0   0
2  -1   4   0   1   0
1 -11  14   0   0   1

2. Use elementary row operations to modify this expanded matrix. (Some notes before I start:

• You probably want to check my arithmetic
• The exact steps I use and the order in which I do them are not the only way to accomplish this step. All that is important is that you use proper elementary row operations correctly and that you end up with the right matrix
• Between each matrix I will describe how I am getting from the one above to the one below. In these descriptions I will refer to row 1 as r1, row 2 as r2 and row 3 as r3.

Replace r2 with r2 + -2*r1:
1   3  -1   1   0   0
0  -7   6   -2  1   0
1 -11  14   0   0   1
Replace r3 with r3 + -r1:
1   3  -1   1   0   0
0  -7   6   -2  1   0
0 -14  15   -1  0   1
Replace r3 with r3 + -2*r2:
1   3  -1   1   0   0
0  -7   6   -2  1   0
0   0   3   3  -2   1
Replace r2 with r2 + -2*r3:
1   3  -1   1   0   0
0  -7   0   -8  5  -2
0   0   3   3  -2   1
Replace r2 with (-1/7)*r2 and replace r3 with (1/3)*r3:
1   3  -1   1    0     0
0   1   0   8/7 -5/7  2/7
0   0   1   1   -2/3  1/3
Replace r1 with r1 + -3*r2:
1   0  -1   -17/7 -15/7 -6/7
0   1   0   8/7   -5/7  2/7
0   0   1   1     -2/3  1/3
Replace r1 with r1 + -r3:
1   0   0   -10/7 -59/21 -11/21
0   1   0   8/7   -5/7    2/7
0   0   1   1     -2/3    1/3

The identity matrix has now "moved" from the right half to the left half.

3. The right half is now the inverse of our original coefficient matrix. So:

-10/7 -59/21 -11/21
 8/7   -5/7    2/7
 1     -2/3    1/3
is the inverse of:
1   3  -1
2  -1   4
1 -11  14

Note: Since the inverse does not have a row of all zeros, then there will be just one solution to this system, the (0, 0, 0) we've know of since the start.

To use this inverse to find a solution to the system, write the system in matrix form and then left-multiply (remember, matrix multiplication is not commutative!) both sides by the inverse matrix. The system, in matrix form is:

  1   3  -1             x             0
  2  -1   4      *      y      =      0
  1 -11  14             z             0
Left-multiply each side by the inverse:
  -10/7 -59/21 -11/21         1   3  -1             x          -10/7 -59/21 -11/21         0
    8/7   -5/7    2/7    *    2  -1   4      *      y      =    8/7   -5/7    2/7    *     0
    1     -2/3    1/3         1 -11  14             z           1     -2/3    1/3          0

On the left side the multiplying first two matrices, if we found the inverse correctly, will result in the identity matrix. And that identity matrix times the variable matrix will result in the variable matrix. On the right side we end up with a 3x1 column matrix with the x value of the solution in the first row, the y value in the second row and the z value in the third row (all zeros as we've known from the start).

P.S. I suggest you multiply the coefficient matrix and its inverse. If you do not get the identity matrix as a result then one or more of the calculations made while we found the inverse must be in error.

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With not just one but both equations already "solved for y", this system is an ideal candidate for using the Substitution Method. The first equation says the y = 2x. Substituting 2x for the y in the other equation we get:
2x = -x + 3
Now we solve for x. Adding x to each side:
3x = 3
Dividing by 3 we get:
x = 1
Now we use this value for x and either of the original equations to find the y value:
y = 2(1)
So y = 2
The solution is the point (1, 2)

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When you're told that triangle ABC is congruent to triangle TUV, you are told more than just the fact that the triangles are congruent.

The letters of the names of triangles represent vertices of the triangle. And usually the order of the letters is unimportant. But when a statement says that two triangles are congruent the order of the letters means something important. The first letters in the names of the triangle are corresponding vertices; the second letters are corresponding vertices and the third letters are corresponding vertices. Since corresponding parts of congruent triangles are congruent we know that if triangle abc is congruent to triangle tuv then:

• Angle A is congruent to angle T
• Angle B is congruent to angle U
• Angle C is congruent to angle V
• Side AB (which is between the first two vertices) is congruent to side TU
• Side BC (which is between the second and third vertices) is congruent to side UV
• Side AC (which is between the first and third vertices) is congruent to side TV

Except for "ABC is congruent to UTV" you should now what the answers are. For "ABC is congruent to UTV" (assuming this means "triangle ABC is congruent to triangle UTV") then this is false because the corresponding parts don't match up. If it had been "triangle BCA is congruent to triangle UVT", then it would be true because the same letters correspond with each other: B and U, C and V and A and T

P.S. Points should be named with capital/upper-case letters. And vertices of a triangle are just special points so they also should be named with capital letters.

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What does "log under root" mean?? You'll have to learn how to read your problems correctly so you can then describe them correctly when you post.

Find out how to read your problem correctly and re-post it in a way we can understand.

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Isn't

already in "y = ..." form? I think you're done!

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Since arctan and tan are inverses of each other, the temptation is to think that the answer would be -5. But it depends on whether the -5 is a number of degrees or a number of radians.

arctan only returns values between values between -90 and 90 degrees or between and radians. So if the -5 is degrees then -5 is the answer because -5 degrees is between -90 and 90. But if the -5 is radians then the answer is not -5 because -5 radians is not between (approximately -1.57) and (approximately 1.57) radians.

If the -5 is radians then what you must do is find an angle that is co-terminal with -5 radians that is between and radians. (Co-terminal angles will always have the same tan's, sin's, etc.)

We can find this co-terminal angle by adding different multiples, possibly including negative multiples, of to -5 radians until we find one and radians.

Since is approximately equal to 6.28, adding to -5 gets us a number approximately equal to 1.28 which is between and radians.

Since we are asked for an exact answer we do not want to use the approximation of 1.27. We use for the exact answer (if the -5 was a number of radians).

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Using the rule:

in both the numerator and denominator we get:


Next we'll use the power of a power rule in the numerator:

Next we'll simplify using the rule :

Now our expression is:

Now we'll use the rule to divide the x's and y's:

Subtracting the fractions:

which simplify to:

This is a simplified expression. Assuming that the "only positive" in your post means "using only positive exponents" then we have a little more work to do. Using the rule we get:

which simplifies to:

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Of course must follow the order of operations (aka PEMDAS or GEMDAS). So we start with the power of a power in the denominator. The rule here is to multiply the exponents:

This is a simplified expression... unless you consider negative exponents "unsimplified". If this is so then we can eliminate the negative exponents quickly or slowly.

Quickly.
The fast way is to understand that negative exponents mean reciprocals. A negative exponent on a factor in the numerator becomes that factor with a positive exponent in the denominator and vice versa. Using this

becomes:

This is the simplified expression with positive exponents. (Note: Exponents only apply to whatever is immediately in front of it! So the exponent of -2 in the denominator applies only to the y, not to the 2! This is why the 2 stays in the denominator.)

Slowly.
The methodical way is to use the rule:

Using this rule on the two negative exponents in:

we get:

Simplifying...

Multiplying the numerator and denominator of the "big" fraction by the lowest common denominator of the "little" fractions, , will eliminate the "little" fractions:

giving us:

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The exponent rule for powers of a power is to multiply the exponents:

The equation now says that one power of a, , is equal to another power of a, . The only way this can be true is if the exponents themselves are equal. So:
4x = 1
Now we just divide by 4:
x = 1/4

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2^x+2^(x+2)=-5y+20
"solve for 2^x in terms of y" means "use algebra to transform the equation so the 2^x is by itself on one side of the equation with the other side being an expression involving y". In other "words":
2^x = expression-with-y-but-not-x
or
expression-with-y-but-not-x = 2^x

So what are we to do with 2^(x+2)? This is hardest part of this problem. The key is to connect a few "dots":

• 2^(x+2) has an exponent that is a sum
• We have a rule for exponents that tells us when exponents should be added:
  
• Most rules in Math, including this one, work in both directions. So even though we usually use the rule above on an expression that matches the pattern of the left side to rewrite it according to the pattern on the right side, we can also use it in the other direction. For example we would usually use this rule to rewrite x^7*x^3 as x^(7+3). But we can also use it to rewrite x^(7+3) as x^7*x^3

So we can use this rule to rewrite 2^(x+2):
2^x + 2^x*2^2 = -5y + 20
Since 2^2 = 4 we get:
2^x + 2^x*4 = -5y + 20
or
2^x + 4*2^x = -5y + 20

And exactly like q + 4q = 5q the left side adds up to:
5*2^x = -5y + 20
Now we just divide by 5:
2^x = -y + 4
And we're finished.

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In order to solve this we need to have the bases of the logarithms match. For this we will use the change of base formula, . We can either change the first log from base x to base 4, the second log from base x to base 4 or change both logs to base whatever-base-we-want. I'm going to change the first log:

The numerator will simplify. When the base and the argument match like this, the log is 1:


Now that the bases of the logs match we can start solving. Let's eliminate the fractions first. Multiplying by :

Using the Distributive Property on the right side:

Simplifying...


The equation is in what is called "quadratic form". If you have trouble recognizing this, then a temporary variable can help:
Let . Substituting this into our equation we get:

This is obviously a quadratic equation. We can solve this. Subtracting 4 (and reordering the terms):

This will not factor so we must use the Quadratic formula:

Simplifying...




which is short for:
or

Of course we are not interested in solutions for q. We want solutions for x. So now we substitute back in for the q:
or
Now we solve for x. The next step is to rewrite these equations in exponential form. Before that, however, let's get decimal approximations for the right sides. Rounded to 3 places we get:
or

Now to exponential form... In general is equivalent to . Using this pattern on our equations we get:
or
Using our calculators to find these powers of 4:
or

Next we check out solutions. This is not optional when solving logarithmic equations like this. You must at least ensure that all bases and all arguments are valid. (Valid bases and arguments are positive and bases cannot be a 1.) We can quickly see that both solutions will make the the bases and arguments of the logs valid. If we had gotten an invalid base or argument, then we would reject that solution.

If you want to complete the check to see if the solutions actually work (this part of the check is optional), then we would need to have logs with bases our calculators "know", base 10 ("log") or base e ("ln"). Using the change of base formula on the original equation:

to convert the logs to base e logs we get:

Then for each solution we replace the x's with that solution and use our calculators to see if it checks. (Note: Because our calculators give us decimal approximations, we may end up with a left side that is very, very close but not exactly equal to the right side. If so then we can assume that the solution checks. Both of these answers check.

We have two correct answers but you say that there is just one answer (which is different from both of the ones we found above). I can only suspect that the reason for this is that there is something wrong with the equation you posted. I hope that you can use the above to help you figure out the solution(s) to the actual equation.

P.S. One you have had some practice with quadratic form equations, you will no longer need a temporary variable. You will be able to see how to go directly from:

to

to

etc.

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The only way two logs of the same base can be equal is if their arguments are equal. So:
5x = 2x + 9
Now solve for x.

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The only way two logs of the same base can be equal is if their arguments are equal. So:
4p-2 = -5p+5
Now solve for p.

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What are you/we supposed to do with this???

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is just an expression and expressions cannot be "solved". But they can be simplified. In this case we can divide the numerator by the denominator.

Synthetic division is designed for division by an expression of the form:
x - n
where "n" is some number. If we are to use synthetic division to divide by x+3 we must imagine (or actually rewrite) it as a subtraction: x - (-3):

-3  |   5   -12   -8
-----       -15   81
       -------------
        5   -27   73

The remainder is in the lower right corner. It is 73. The rest of the bottom row is the quotient. The "5 -27" translates into 5x-27. So after dividing:

we get:

(Note what happens with the remainder. We handle in just the same way as in the "good old days" before variables: We make a fraction with the remainder on top and the divisor on the bottom!)

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a. To find the zeros of a polynomial of degree 2 or more we factor the polynomial. (Although in the special case of Quadratics we can use the Quadratic Formula instead of factoring.) The greatest common factor is 1 (which we rarely factor out). There are too many terms for any of the factoring patterns or for trinomial factoring. I don't see how to factor the polynomial by grouping so we are left with factoring by trial and error of the possible rational roots.

The possible rational roots of a polynomial are all the ratios, positive and negative, that can be formed using a factor of the constant term (at the end) in the numerator over a factor of the leading coefficient (at the front). The contant term in our polynomial is 5. (Actually it is -5 but since we are going to use all positive and negative ratios it doesn't matter if we use the "-".) The factors of 5 are just 1 and 5. The leading coefficient (in front of ) is 1 which just has 1's for factors. This makes the possible rational roots of our polynomial:
+1/1 or +5/1
which reduce to:
+1 or +5

Now we see if any of these actually are roots/zeros of the polynomial. Since powers of 1 are very simple we can easily check that mentally. But since the problem tells you to use synthetic division, we will do so:

1  |   1   -4   8   -5
----        1  -3    5
      -----------------
       1   -3   5    0

The remainder, in the lower right corner is zero. This means that (x-1) is a factor and that 1 is a root/zero of the polynomial. Not only that but the rest of the bottom row tells us what the other factor will be. "1 -3 5" translates into . So f(x) in (partially) factored form is:


The remaining zeros of the polynomial will be zeros of the second factor. That factor will not factor further, no matter which factoring techniques we use! But it is quadratic so we can use the formula:

Simplifying...




which is short for:
or
With the negative inside the square root, these roots/zeros will be complex numbers. So we should express them in standard a + bi form:
or
or
or
or
or

These two, plus the zero of 1 we found earlier, are the three zeros of your polynomial function.

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• What are you/we supposed to do with that expression?
• Put parentheses around numerators and denominators, especially if they have multiple terms. What you posted meant:
  
  which I suspect is not the expression you intended.
• The expression has nothing to do with logarithms. Please post questions under a relevant category.

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What are you/we supposed to do with that equation?

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Since all multiples of are special angles and since the constant part of the argument, , is already a multiple of , it will probably be easiest if we choose values for x that would make the other term in the argument, , also be a multiple of . By doing so we would then be able to add the fractions and end up with an argument that is also a multiple of .

Perhaps you can already tell what values x should have that would turn into a multiple of . If not, then just pick a multiple of , set equal to it and solve for x. Let's be easy on ourselves and choose itself:

Multiplying both sides by 6 we get:

Dividing by 3 we get:


Perhaps you can now see that if x is any multiple of , then would end up being some multiple of . I'll get you started on a table of values:

  x    (1/2)x   (1/2)x+5pi/6  sin((1/2)x+5pi/6)   2sin((1/2)x+5pi/6)   f(x)
pi/3    pi/6      6pi/6 = pi        0                  0                0
-pi/3   -pi/6   4pi/6 = 2pi/3     sqrt(3)/2            sqrt(3)         sqrt(3)
  0       0         5pi/6          1/2                 1                1

I'll leave it up to you to find two more.

P.S. We did not have to make x be a value that made the argument of sin be a multiple of pi/6. We just did that because it made things easier. Any x that would make (1/2)x+5pi/6 turn out to be any of the special angles would work.