# Questions on Algebra: Trigonometry answered by real tutors!

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Question 994584: cos theta = - sqrt55/8 and 90 degrees < theta < 180 degrees. Find sin 2theta
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cos theta = - sqrt55/8 and 90 degrees < theta < 180 degrees. Find sin 2theta.
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Probably,  the condition is read as   = -.

OK,  if so,  then

= = = = .

The sign is  "plus"  since the angle    is in the second quadrant.

Then   = = - = -.

Question 994566: Prove that:
2cos^3 θ-cos θ/sinθ cos^2 θ-sin^3 θ= cot θ

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Can't. It is only true at odd multiples of pi/2. Without the lead coefficient of 2 in the numerator, then it works. Was that a typo?

John

My calculator said it, I believe it, that settles it

Question 994411: The width of a rectangular is 6 ft less than 4 times the length. Write a model for the width W in terms of the length L.

Question 994358: prove that : 1 - sin theta / 1 + sin theta = ( sec theta- tan theta)2
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(For easier typing, we will let theta = x)
If we multiply the numerator and denominator by 1 - sin(x), and simplify we get:
(1 - 2sin(x) + sin^2(x))/cos^2(x) = (1-2sin(x))/cos^2(x) + tan^2(x)
Simplifying the first term gives
sec^2(x) - 2sec(x)tan(x) + tan^2(x) = (sec(x) - tan(x))^2

Question 993812: You are supposed to show the steps it takes to get the answer:
(sec²x - 6 tan x + 7)/ (sec²x - 5) = (tan x - 4)/ (tan x + 2)

Found 2 solutions by KMST, Alan3354:
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I assume you want to find the value(s) of that is/are solutions to the equation above.
Solutions must make the expressions above exist.
The functions and must be defined,
and the denominators must not be zero:
<--><--> and
<-->
So far, we know that the solution is such that
.
For any other value of , we try to simplify and solve.
There may be better ways, but this is what I came up with:

<--><--><--><--><--><--> .
The last equation is an identity.
So, the answer is any such that .

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Question 993808: You are supposed to show the steps it takes to get the answer:
(1- 3cosx - 4 cos²x)/ (sin²x) = (1- 4cosx)/ (1- cosx)

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I don't think I am.

Question 994020: from 1990 to 1998 the manufacturer's shipments for audio cassettes A (in millions) and compact discs C (in millions) can be modeled by the equations:
A=-31.8t+322
c=428.8t+110
where t is the number of years since 1990. During what year did the number of compact discs shipped surpass the number of audio cassettes shipped?

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The point where the two graphs meet is the year, if one looks at t.
-31.8t +322=428.8t + 110
460.6 t= 212
t=0.460. This would be in mid-1990. I wonder if the numbers were copied correctly, but with those equations, that is the value.

Question 993981: Find the solutions of the equation that are in the interval [0, 2pi;). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2-2sint=2sqr(3)cost

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I did this one yesterday.

Question 993889: A farmer has a triangular shaped field, FLD. DF-350m, FL= 240m and DL=400m.The farmer wants to put up a fence f, and has exactly enough fence such that angle FXL=71 degrees. Calculate the length of fence f.
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A farmer has a triangular shaped field, FLD. DF-350m, FL= 240m and DL=400m.The farmer wants to put up a fence f, and has exactly enough fence such that angle FXL=71 degrees. Calculate the length of fence f.
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Where is X ?

Question 993895: The diagram shows the location of airports P, Q and R. Airport Q is 130 km from Airport P on a bearing of 111 degrees. Airport R is 75 km from Airport Q on a bearing of 225 degrees.
a. Calculate angle PQR
b. Calculate the distance from Airport R to Airport P, giving your answer to the nearest whole number.
c. Calculate angle RPQ
d. Calculate the bearing of Airport P from Airport R.

Question 993896: The diagram shows the location of airports P, Q and R. Airport Q is 130 km from Airport P on a bearing of 115 degrees. Airport R is 75 km from Airport Q on a bearing of 225 degrees.
a. Calculate angle PQR
b. Calculate the distance from Airport R to Airport P, giving your answer to the nearest whole number.
c. Calculate angle RPQ
d. Calculate the bearing of Airport P from Airport R.

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What diagram?

Question 993849: Sin3A-SIN2A equal to zero then what is the value of A

Found 2 solutions by ikleyn, Alan3354:
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.
If your equation is  sin(3A) = sin(2A)  then

3A + 2A = ,

5A = ,

A = ,       or

A = = 36°.

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O!  There is one more solution:  A = 0,  but it is not so interesting.

A = 180° is the solution also, by the way.

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Sin3A-SIN2A equal to zero then what is the value of A
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Is it sin^3(A) - sin^2(A) ?
or sin(2A) - sin(3A) ?

Question 993872: A farmer has a triangular shaped field, FLD. DF-350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.
Found 2 solutions by ikleyn, Alan3354:
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.
A farmer has a triangular shaped field, FLD. DF-350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.
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There are  3  angles.  Angle  F  is opposite the longest side.
Use the Cosine Law:
400^2 = 350^2 + 240^2 - 2*350*240*cos(F).

I got  m(LF) =~ 83° (approximately).

For the Cosine Law see,  for example,  the lesson  Proof of the Law of Cosines revisited  in this site.

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A farmer has a triangular shaped field, FLD. DF-350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.
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There are 3 angles. Angle F is opposite the longest side.
Use the Cosine Law:
400^2 = 350^2 + 240^2 - 350*240*cos(F)

Question 993803: You are supposed to show the steps it takes to get the answer:
sec²∅ + csc²∅ = sec²∅csc²∅
(∅= theta)

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sec^2 + csc^2 = sec^2*csc^2
Multiply thru by sin^2*cos^2
----
sin^2 + cos^2 = 1
I'm convinced.
==========================
PS If anyone can post an Identity proof that shows a difference when working on one side only I would like to see it.

Question 993791: Find sin2b if sin b = 7/25
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Find sin2b if sin b = 7/25
-----------------------------------

Is my understanding correct that you want to find  sin(2b)?

If so,  use the formula   sin(2b) = 2*sin(b)*cos(b).

Do you know this formula?

To complete calculations,  you need to know  cos(b)  too.
You can calculate it via  sin(b):

cos(b) = = = = = = .

Now you are ready to calculate

sin(2b) = 2*sin(b)*cos(b) = .. = . . .

Question 993815: You are supposed to show the steps it takes to get the answer:
(sin³A + cos³A)/ (sin A + cos A)= 1 - sin A cos A

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.
Use this equality:

= ..

It is the standard factorization and it is described in any systematic Algebra textbook.  See,  for example,  the lesson  The cube of the sum formula  in this site.

Therefore,   = .

Now substitute  x = sin(A),  y = cos(A)  into this formula.  You will have

= = .

We took into account that   = .

Question 993701: I'm in pre calculus and I need help with solving a problem. It is y=cot(x-pi/4) I need to find the period, frequency, phase shift, vertical shift, amplitude, and the asymptotes.
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below is the graph of the equation of y = cot(x) and y = cot(x - pi/4).

the graph of y = cot(x) is in orange.
the graph of y = cot(x - pi/4) is in black.

you will see that the graph of y = cot(x - pi/4) is shifted to the right of the graph of y = cot(x) by pi/4 radians.

the normal interval of the cotangent function is 180 degrees, or pi.

this means the graph repeats every 180 degrees, or every pi radians.

interval shown for y = cot(x) is 0 to pi.
interval shown for y = cot(x - pi/4) is pi/4 to 5pi/4.

the general form of a sinusoidal type wave is:

y = a * sin(b * (x-c)) + d or:
y = a * cos(b * (x-c)) + d or:
y = a * tan(b * (x-c)) + d or:
y = a * cot(b * (x-c)) + d or:

a is the amplitude
b is the frequency
c is the horizontal shift
d is the vertical shift.

sin and cos are sinusoidal type of wave with a period of 360 degrees because their pattern repeats every 360 degrees.

tan and cot are a sinusoidal type of wave with a period of 180 degrees because their pattern repeats every 180 degrees.

you want to find the period, frequency, phase shift, vertical shift, amplitude, and the asymptotes for the equation of y = cot(x - pi/4).

the period is 180 degrees, or pi radians.
the frequency is 1 because it is now shown.
the phase shift is the same as the horizontal shift which is pi/4.
the vertical shift is 0 because it is not shown..
the amplitude is 1 because it is not shown.
the asymptotes are at pi/4 and 5pi/4.

Question 993709: If cot 17 equals 3.2709 find tan 73
Found 3 solutions by MathTherapy, ikleyn, Boreal:
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If cot 17 equals 3.2709 find tan 73
cot and tan are CO-FUNCTIONS, so
Therefore,  _________
Thus,



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.
tan(73°) = .

sin(73°) = cos(90°-73°) = cos(17°)     (reduction formula).

cos(73°) = sin(90°-73°) = sin(17°).       - - - " - - -

Therefore,  tan(73°) = = = cot(17°) = 3.2709.

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cot 17=tan 73.
It is 3.2709.
They are reciprocals, just like sin 60= cos 30.

Question 993593: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?

Question 993591: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?

Question 993598: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
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Posted too many times.

Question 993602: 1.the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
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Posted too many times.

Question 993604: 1the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
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Posted too many times.

Question 993592: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
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tan(48) = y/x

tan(32) = y/(x+5)

solve for y in both equation to get:

y = x * tan(48)

y = (x+5) * tan(32)

since they're both equal to y, the expressions on the right side of each equation must be equal to each other, so:

x * tan(48) = (x + 5) * tan(32)

simplify the right side of the equation to get:

x * tan(48) = x * tan(32) + 5 * tan(32)

subtract x * tan(32) from both sides of the equation to get:

x * tan(48) - x * tan(32) = 5 * tan(32)

simplify the left side of the equation to get:

x * (tan(48) - tan(32)) = 5 * tan(32)

divide both sides of the equation by (tan(48) - tan(32)) to get:

x = (5 * tan(32)) / (tan(48) - tan(32))

solve for x to get:

x = 6.432096215

x + 5 must therefore be equal to 11.432096215

now that you know the value of x and x + 5, you can use that information to solve for y.

it should be the same in both cases.

if y = x * tan(48), then y is equal to 6.432096215 * tan(48) which is equal to 7.143566553.

if y = (x + 5) * tan(32), y is equal to 11.43209621 * tan(32) which is equal to 7.143566553.

y is the same in both cases, as it should be.

your solution is that the length of the longer shadow of the pole is 6.432096215 + 5 = 11.432096215

Question 993633: If sin A=3/5 then find (tan A+sec A)^2
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If sin A=3/5 then find (tan A+sec A)^2
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Since sin = y/r, y = 3 and r = 5
---
Then, x = sqrt[5^2-3^2] = sqrt(16) = 4
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So, tan(A) = y/x = 3/4 and sec(A) = r/x = 5/4
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Ans: (tan(A)+sec(A))^2 = (3/4+5/4)^2 = 2^2 = 4
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Cheers,
Stan H.
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Question 993622: Please help me out with this problem! Solve for all solutions of t on the interval [0,2pi)
2-2sin(t)=2sqrt( 3 )cos(t)

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Please help me out with this problem! Solve for all solutions of t on the interval [0,2pi)
2-2sin(t)=2sqrt(3)cos(t)

Square both sides.

Sub 1 - sin^2 for cos^2
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(2sin + 1)*(sin - 1) = 0
----
sin(t) = 1
t = pi/2
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sin(t) = -1/2
t = 7pi/6, 11pi/6

Question 993625: Please help!! Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0

Found 2 solutions by stanbon, Alan3354:
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Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0
----
sin(t) = [8+-sqrt(64-4)]/2
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sin(t) = [8+-sqrt(60)]/2
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sin(t) = 8+-sqrt(15)
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Comment:: Both values for sin(t) are out of the range of the sin function.
---
Ans: No solution.
Cheers,
Stan H.
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Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0
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Sub x for sin(t)

 Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=60 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 7.87298334620742, 0.127016653792583. Here's your graph:

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sin(t) =~ 0.12701665
t =~ 7º 20'
t =~ 172º 40'

Question 993471: Find all solutions of the equation in the interval [0,2pi)
2cos^2x-cosx=1

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Let

Then

Solve the quadratic and the substitute back. Find all values of that have cosine function values of the roots of the equation.

John

My calculator said it, I believe it, that settles it

Question 993436: We are supposed to show the process that gets you to the answer (1) in the problem
csc²∅- cos²∅csc²∅= 1
(∅= theta)

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Hi there,
csc²∅- cos²∅csc²∅= 1
[csc²∅ = 1/sin^2∅]
.......
csc²∅- cos²∅csc²∅= 1
1/sin^2∅- cos²∅ 1/sin^2∅ = 1
1 - cos²∅/sin^2∅ = 1
[1 - cos²∅ = sin^2∅]
sin^2∅/sin^2∅ = 1
Hope this helps :-)

Question 993170: A circle has a radius of 5ft. Find the length s of the arc intercepted by a central angle of 140°

Do not round any intermediate computations, and round your answer to the nearest tenth.

Question 992753: Given that cotx=-5/6 and x is in quadrant 2, find sin2x, cos2x,tan2x
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.
Given that cotx=-5/6 and x is in quadrant 2, find sin2x, cos2x,tan2x
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cot(x) =     --->      = ---> = .     --->

= ..         (1)

+ = 1,         (2)

Now substitute  (1)  into  (2).  You will get

. = 1,     or

= = .

Hence,   sin(x) = ,   cos(x) = -     (taking into account that  x  is in the second quadrant).

Now,  when you know  sin(x)  and  cos(x)  values,  you can calculate the remaining functions.

Use the trigonometry formulas

sin(2x) = 2sin(x)*cos(x), = . . . ,

cos(2x) = - = . . . ,

and then  tan(2x) = = . . . .

Complete these calculations yourself.

Good luck!

Question 992819: Given that tan θ − cot θ = 2,
find the possible values of tan θ, giving your answers in an exact form.
Hence solve the equation
tan θ − cot θ = 2
giving all values of θ between 0 and 360
I'm very confused how the questions are different? I really don't understand what the first question is asking me to do.

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.
Let  x  be  tan(x):   x = tan(x).

Then  cot(x) = ,  and you have an equation

x - = 2.

Simplify and solve it:

- - = ,

= .

The root is   x = 1.

Thus  tan(x) = 1.

x = = 45°  and/or  x = = = 225°.

Question 992746: Given that cotx=3/5 and x is in quadrant 2, find sin2x, cos2x,tan2x
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.
Hey, cot(x) is negative in the second quadrant.

Do you know it?

Question 992754: Given that sinx=3/5 and it is in quadrant 1. Find sin2x+cos2x
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Hi there,
If sin x = 3/5
cos x = 4/5
tan x = 3/4
..........
sin 2x = 2 sin x cos x
sin 2x = 2 (3/5)(4/5)
sin 2x = 24/25
..........
cos 2x = 2 cos^2 x - 1
cos 2x = 2(4/5)^2 - 1
cos 2x = 2(16/25) - 1
cos 2x = 32/25 - 1
cos 2x = 32/25 - 25/25
cos 2x = 7/25
..........
Hope this helps :-)

Question 992757: if tan theta = 1 / root 7, show that (cosec ^2 theta - sec ^2 theta )/cosec^2 theta +sec^2 theta )=3/4
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Hi there,
if tan theta = 1 / root 7
cosec theta = root 8/1
sec theta = root 8/root 7
...........
(cosec ^2 theta - sec ^2 theta )/cosec^2 theta +sec^2 theta )
(root 8/1)^2 - (root 8/root 7)^2/ (root 8/1)^2 + (root 8/root 7)^2
(8/1 - 8/7) / (8/1 + 8/7)
((56 -8)/7 / ((56 + 8)/7
[48/7 / 64/7]
[ 48 x 7]/ [7 x 64)
Cancel 7's top and bottom
=48/64
Divide top and bottom by 16
= 3/4
Hope this helps :-)

Question 991736: Find all solutions of csc x = -2, where 0 ≤ x ≤ 2π. Can you explain step by step how you got the answer.
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Find all solutions of csc x = -2, where 0 ≤ x ≤ 2π.
-----
If csc(x) = -2, sin(x) = 1/csc(x) = -1/2
------
If sin = -1/2, the reference angle is pi/6
------
Since sin is negative in QIII and QIV
x = pi+(pi/6) = (7/6)pi
or x = (2pi)-(pi/6) = (11/6)pi
-------------------
Cheers,
Stan H.
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Question 992125: Y=1/2x x is great than or equal to 0 x>/=0
Graph and calculate the values for the six trigonometric functions of the angle theta given in standard position, if the terminal side of theta lies on the given line.
Sin=
Cos=
Tan=
Cose=
Sec=
CoTan=

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Y=1/2x x is great than or equal to 0 x>/=0
If x = 1, y = 1/2
Then r = sqrt[1+(1/2)^2] = sqrt(3/4) = sqrt(3)/2
----------------

Graph and calculate the values for the six trigonometric functions of the angle theta given in standard position, if the terminal side of theta lies on the given line.
Sin= y/r = 1/sqrt(3)
-----
Cos= x/r = 2/sqrt(3)
------------
Tan= y/x = 1/2
---------------------
Invert sin, cos, tan to get csc, sec, and cotan
----------
Cheers,
Stan H.
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Cose=
Sec=
CoTan=

Question 992448: solve for x E[0,2pi] :2sin^2x-sinx=0
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solve for x E[0,2pi] :
========================
2sin^2x-sinx=0
Factor::
sin(x)[2sinx-1] = 0
----
sin(x) = 0 or sin(x) = 1/2
---
x = 0 or x = pi/6 or x = (5/6)pi
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cheers,
Stan H.
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Question 992389: Given sin t = 1/5, sec t<0, find cos (t) and cot (t).
I do not know how to find out if any of the functions (cos, cot, or sin) are negative or positive.
This is what I have:
y=1,, r=5, and x=2√6
cos⁡(t)=(2√6)/5, cot⁡(t)=1/(2√6)
I don't know where to go from here