Solve for x in terms of pi if 2sin^2x - cosx - 1 =0 for 0 < x<pi

Question

Solving for x in terms of pi

Answer 1

2sin²x - cosx - 1 = 0

 

2(1 - cos²x) - cosx - 1 = 0

 

-2cos²x - cosx + 1 = 0

 

2cos²x + cosx - 1 = 0

 

(2cosx - 1)(cosx + 1) = 0

 

cosx = 1/2 or cosx = -1

 

If cosx = 1/2 and 0 < x < π, then x = π/3

 

There are no solutions to the equation cosx = -1 under the restriction  0 < x < π.

 

So, x = π/3.