SOLUTION: Please help me solve this equation. I have a huge trig test tomorrow and I have no idea what I am doing... csc^2(2x)-csc(2x)=2

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Question 1007411: Please help me solve this equation. I have a huge trig test tomorrow and I have no idea what I am doing... csc^2(2x)-csc(2x)=2
Found 3 solutions by ikleyn, lwsshak3, MathTherapy:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
 -  = .

Introduce new variable y = csc(2x).

Then your equation takes the form 

 = .

Factor it:

(y+1)*(y-2) = 0

The solutions are y = -1 and y = 2.

Now you need to solve two equations:

1) csc(2x) = -1 ----->  = -1 -----> sin(2x) = -1 -----> 2x = 270° -----> x = 135°.

2) csc(2x) = 2 ----->  = 2 -----> sin(2x) =  -----> 2x = 30° and 2x = 150° -----> x = 15° and x = 75°.

Answer. The solutions are x = 15°, x = 75° and x = 135°.


Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
csc^2(2x)-csc(2x)=2
let x=csc(2x)
x^2-x-2=0
(x+2)(x-1)=0
x=-2=csc(2x) (reject)
or x=1=csc (2x)
2x=π/2
x=π/4

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Please help me solve this equation. I have a huge trig test tomorrow and I have no idea what I am doing... csc^2(2x)-csc(2x)=2

------- Converting from csc to sin
---------- Multiplying by LCD,

[2 sin (2x) - 1][sin (2x) + 1] = 0 ------ Factoring trinomial
2 sin (2x) – 1 = 0 AND sin (2x) + 1 = 0
2 sin (2x) = 1 AND sin (2x) = 0 - 1
AND sin (2x) = - 1
AND
30 = 2x AND 270 = 2x
(Q I) AND (Q II)
(Q II)
No INTERVAL was stated (the likes of: ) so these are the 3 solutions
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