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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