How do I solve for the value of x in degrees given (2sinx - 1)(2cos²x - 1) = 0?

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Answer 1 · 2018-02-21T23:36:21

#x=30^@, 45^@, 135^@, 150^@, 225^@, 315^@#

We start by splitting up the problem into two:

#(2 sin x - 1)(2 cos^2 x - 1) =0#
#2 sin x - 1 = 0#
#2 cos^2 x - 1 = 0#

We'll tackle each of these individually. I'm assuming that #0 <= x <= 360#.

#2 sin x - 1 = 0#
#2 sin x = 1#
#sin x = 1/2# (have you memorized your unit circle? :) )
#x = 30^@, 150^@#

#2 cos^2 x - 1 = 0#
double angle identity!
#cos(2x) = 0#
#x = 45^@, 135^@, 225^@, 315^@#

It may seem surprising that there are so many answers, but a quick graph on Mathematica confirms our solution.

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