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
Nghi N.
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#

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