This is a little it hard... I am wondering if others have a better way to solve it.
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2sin2x−cosx=1
4sinxcosx = 1 + cosx
16sin^2xcos^2x = 1 + 2cosx + cos^2x
16(1 - cos^2x)cos^2x = 1 + 2cosx + cos^2x
16cos^2x - 16cos^4x = 1 + 2cosx + cos^2x
16cos^4x - 15cos^2x + 2cosx + 1 = 0
let cosx = Y
16Y^4 - 15Y^2 + 2Y + 1 = 0
(Y + 1)(16Y^3 - 16Y^2 + Y + 1) = 0
there are four values for Y, -1, -0.2, 0.367, 0.836.
Then you solve for Y.
For example, Y = -1 means cosx = -1, x = π.
I will leave the rests to you. (Sorry, this seems to be an ugly way...)
