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

5π6;π6;π2

Explanation:

2sin2xcos(x+π2)1=0
Note. cos(x+π2)=sinx
2sin2xsinx1=0
Solve this quadratic equation foe sin x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
sin x = 1, and sinx=ca=12
a. sin x = 1 --> x=π2
b. sinx=12
Trig table and unit circle give 2 solutions for x:
x=π6 and sinx=5π6
Answers for (-pi, pi):
5π6;π6;π2