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.

1 Answer
Jun 18, 2018

-(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