SOLUTION: I'm attempting to do corrections on a test, but these three questions are giving me trouble! I've tried several different things, and I keep coming up with the same (wrong!) answer

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Question 170144: I'm attempting to do corrections on a test, but these three questions are giving me trouble! I've tried several different things, and I keep coming up with the same (wrong!) answer!
1) If x is an angle in quadrant 4 and Cot%28x%29=+-7%2F24, find the value of Sin1%2F2x.
2) What is the solution set of the equation
Sin%282x%29+-+Cos%5E2x%2B1+ = +Sin%5E2x%2B+Sin%28x%29+ in the interval 0%3C=x%3C2pi?
3) Find the solution set of 6Sin%28x%29+%2B+11=+2Csc%28x%29 over the domain 0%3Cx%3C360°.

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(48558) About Me  (Show Source):
You can put this solution on YOUR website!
1) If x is an angle in quadrant 4 and cotx= -7/24, find the value of
sin[(1/2)x].
If cotx = -7/24, x = 7 and y = -24
Then r = sqrt(7^2 + 24^2) = 25
------------
sin[(1/2)x] = sqrt[(1-cos(x))/2]
So, sin[(1/2)x] = sqrt[(1-(7/25))/2] = sqrt[9/25] = 3/5
=============================================================
2) What is the solution set of the equation sin(2x)-cos^2(x+1) = sin^2(x)+sinx in the interval 0 Comment: That is a mess to analyze. Graph the left side and the right side
separagely and see where they intersect under the condition the 0 =============================================================
3) Find the solution set of 6sinx + 11= 2cscx over the domain 0 Multiply thru by sin(x) to get:
6sin^2(x) + 11sin(x) - 2 = 0
---
Let w = sin(x)
Substitute to get:
6w^2 + 11w - 2 = 0
6w^2 +12w - w -2= 0
6w(w+2) -(w+2) = 0
(w+2)(6w-1) = 0
w = -2 or w = 1/6
-----
Solve for x:
sin(x) = -2 or sin(x)= 1/6
sin(x) cannot be -2.
If sin(x) = 1/6, x = 9.594.. degrees or 170.41 degrees
========================================================
Cheers,
Stan H.





Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
Stanbon's solution is correct but he does not do problem 2 and he does not explain why we know to take the positive square root in 1.
Edwin's solution: 
 
1) If x is an angle in quadrant 4 and Cot%28x%29=+-7%2F24, find the value of Sin1%2F2x.

Since this involves drawing a graph in which x represents
the horizontal axis, not an angle,  I will temporarily change
x to alpha%29 to avoid a conflict of letters.  Change
the problem to read this way:

1) If alpha is an angle in quadrant 4 and Cot%28alpha%29=+-7%2F24, find the value of Sin1%2F2alpha.

We must use the identity:

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%281-Cos%28alpha%29%29%2F2%29

However we do not know Cos%28alpha%29

So we must first draw the picture of the angle alpha:

We know that Cot%28alpha%29 is by definition x%2Fy, we
can draw the angle in the 4th quadrant with referent angle
is inside a triangle whose horizontal side x is taken
to be the numerator of -7%2F24, considered positive because
it goes right of the y-axis, and whose vertical side y is taken as 
the denominator -24, taken negative because it goes down
below the x-axis: 

drawing%28400%2C400%2C-30%2C30%2C-30%2C30%2C+graph%28400%2C400%2C-30%2C30%2C-30%2C30%29%2C%0D%0A%0D%0Aline%280%2C0%2C7%2C-24%29%2Cline%287%2C0%2C7%2C-24%29%2Cline%280%2C0%2C7%2C0%29%2C+locate%288%2C-10%2C%27y=-24%27%29%2C%0D%0Alocate%282%2C4%2C%27x=7%27%29+%29 

Next we calculate r by the Pythagorean theorem:

r%5E2=x%5E2%2By%5E2
r%5E2=%287%29%5E2%2B%28-24%29%5E2
r%5E2=49%2B576
r%5E2=625
r=sqrt%28625%29
r=25

So we label the slanted line segment r=25. 

drawing%28400%2C400%2C-30%2C30%2C-30%2C30%2C+graph%28400%2C400%2C-30%2C30%2C-30%2C30%29%2C%0D%0Alocate%28.5%2C-18%2C%27r=25%27%29%2C%0D%0Aline%280%2C0%2C7%2C-24%29%2Cline%287%2C0%2C7%2C-24%29%2Cline%280%2C0%2C7%2C0%29%2C+locate%288%2C-10%2C%27y=-24%27%29%2C%0D%0Alocate%282%2C4%2C%27x=7%27%29+%29 

Now we can find Cos%28alpha%29

Cos%28alpha%29=x%2Fr=7%2F25

So we substitute 7%2F25 for Cos%28alpha%29 in

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%281-Cos%28alpha%29%29%2F2%29

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%281-7%2F25%29%2F2%29

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%2825%2F25-7%2F25%29%2F2%29

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%2818%2F25%29%2F2%29

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%28%2818%2F25%29%281%2F2%29%29

Sin1%2F2alpha%27=%27%27%27%2B-sqrt%289%2F25%29

Sin1%2F2alpha%27=%27±3%2F5

Next we must decide whether this is positive or negative:

Since alpha is is the 4th quadrant, then 

270%3Calpha%3C360° so multiplying that through by 1%2F2

135%3C1%2F2alpha%3C180° 

The means 1%2F2alpha is in quadrant 2.  Since
the sine is positive in the 2nd quadrant, the final answer
is 

Sin1%2F2alpha=3%2F5 

And of course now that we have the answer we can change
alpha back to x:

Sin1%2F2x=3%2F5


2) What is the solution set of the equation

 Sin%282x%29+-+Cos%5E2x%2B1+ = +Sin%5E2x%2B+Sin%28x%29+ in the interval 0%3C=x%3C2pi?

Use the identity Cos%5E2x=+1-Sin%5E2x to replace Cos%5E2x on the
left side:

 Sin%282x%29+-+%281-Sin%5E2x%29+%2B1+ = +Sin%5E2x%2B+Sin%28x%29+

 Sin%282x%29+-+1%2BSin%5E2x+%2B1+ = +Sin%5E2x%2B+Sin%28x%29+

Sin%282x%29+%2BSin%5E2x+ = +Sin%5E2x%2B+Sin%28x%29+

Sin%282x%29++ = +Sin%28x%29+

Sin%282x%29+-+Sin%28x%29 = 0

Now use identity Sin%282x%29+=+2%2ASin%28x%29Cos%28x%29 to
replace Sin%282x%29

2%2ASin%28x%29Cos%28x%29+-+Sin%28x%29 = 0

Factor out Sin%28x%29

Sin%28x%29%282Cos%28x%29+-+1%29 = 0

Use the zero-factor principle:

matrix%285%2C4%2CSin%28x%29=0%2C+++%27%27%2C+%27%27%2C+++2Cos%28x%29-1=0%2C%0D%0A+++++++++++x+=+0%2C++++++%27%27%2C+%27%27%2C++++2Cos%28x%29=1%2C+++++%0D%0A+++++++++++x+=+pi%2C+++++%27%27%2C+%27%27%2C+++++Cos%28x%29=1%2F2%2C%0D%0A+++++++++++++%27%27%2C+++++++%27%27%2C+%27%27%2C++++++x=pi%2F3%2C%0D%0A+++++++++++++%27%27%2C+++++++%27%27%2C+%27%27%2C++++++x=5pi%2F3+%29

3) Find the solution set of 6Sin%28x%29+%2B+11=+2Csc%28x%29 over the domain 0%3Cx%3C360°.

6Sin%28x%29+%2B+11=+2Csc%28x%29

Use identity Csc%28x%29=1%2FSin%28x%29


6Sin%28x%29+%2B+11=+2%281%2FSin%28x%29%29

Multiply through by Sin%28x%29

6Sin%5E2x%2B11Sin%28x%29=2

6Sin%5Ex%2B11Sin%28x%29-2=0

%286Sin%28x%29-1%29%28Sin%28x%29%2B2%29=0

Use the zero-factor principle:
 
matrix%286%2C4%2C%0D%0A++++6Sin%28x%29-1=0%2C++%27%27%2C++%27%27%2C++Sin%28x%29%2B2=0%2C%0D%0A+++++6Sin%28x%29=1%2C+++%27%27%2C++%27%27%2C+++Sin%28x%29=-2%2C%0D%0A++++++Sin%28x%29=1%2F6%2C+%27%27%2C++%27%27%2C+++ignore%2C++%0D%0A++++++%27x=9.594%B0%27%2C+%27%27%2C+++%27%27%2C+++since%2C%0D%0A++++++%27x=180%B0-9.594%B0%27%2C+%27%27%2C+%27%27%2C+abs%28Sin%28x%29%29%3C1%2C+++%0D%0A++++++%27x=170.406%B0%27%2C+%27%27%2C+%27%27%2C+%27%27+%29+

Edwin