Question

How would I solve this? 2sin^2 x - cos(x + pi/2) - 1 = 0 for -pi < x < pi

I’m really sorry I’m not too sure how to do all the symbols.
Thanks to anyone that can help.

Answer 1

`-(5pi)/6; -pi/6; pi/2`

Explanation:

`2sin^2 x - cos (x + pi/2) - 1 = 0`
Note. `cos (x + pi/2) = sin x`
`2sin^2 x - sin x - 1 = 0`
Solve this quadratic equation foe sin x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
sin x = 1, and `sin x = c/a = -1/2`
a. sin x = 1 --> `x = pi/2`
b. `sin x = - 1/2`
Trig table and unit circle give 2 solutions for x:
`x = - pi/6` and `sin x = - (5pi)/6`
Answers for (-pi, pi):
`(-5pi)/6; - pi/6; pi/2`