For solutions between 0 and 2pi, solve 2cot^(2)x = 5 - cosec x

Question

I don't find an answer for this question matching the solution, please help me out! I need to get exact values for this, solved by solving the equation. 

The range is 0 to 2π

Solve 2cot²x = 5 - cosec x

Answer 1

Solve

 

2 cot² x = 5 - csc x.

 

Write in terms of sines and cosines only. Recall cot x = (cos x)/(sin x) and csc x = 1/(sin x). Plug these in.

 

2 ((cos x)/(sin x))² = 5 - 1/(sin x)

 

(2 cos ² x)/(sin² x) = 5 - 1/(sin x)

 

2 cos ² x = 5 sin² x - sin x    (after clearing the denominators and multiplying both sides by sin² x)

 

2(1 - sin² x) = 5 sin² x - sin x    (use the identity cos² x = 1 - sin² x to rewrite using sin x only)

 

2 - 2 sin² x = 5 sin² x - sin x

 

0 = 7 sin² x - sin x - 2

 

This is a quadratic in sin x. You will need to use the quadratic formula to solve this, as it cannot be factored. This probably means that the solutions for x may not be writable in exact form.

 

I recommend you try to solve it this way and see what happens. If you cannot get an exact answer, it's probably because the problem or the answer has an error. You should check with your teacher to be certain. (I could not get exact solutions to this myself, which is why I suggest you recheck the problem.)