2 cos x - sin x + 2 cos x sin x = 1
2 cos x - sin x = 1 - 2 cos x sin x
Squaring, possibly introducing extraneous roots,
4 cos^2 x + sin ^2 x - 4 cos x sin x = 1 + 4 cos^2 x sin ^2 x -4 cos x sin x
The cross term cancels, that's fortunate. Let's turn everything to cosine and let c=cos ^2 x.
4 c + 1 - c = 1 + 4 c (1 - c )
-c = -4c^2
c(4c - 1) = 0
cos^2 x = c = 0 or cos^2 x = c= 1/4
cos x = 0 or cos x = pm 1/2
In the requested range, x=pi/2, x={3pi}/2 from the first, x =pi/3, {2pi}/3, {4pi}/3, {5pi}/3 from the second.
We squared an equation so we need to check these.
2 cos (pi/2) - sin (pi/2) + 2 cos (pi/2) sin (pi/2) = -1 quad NOPE
2 cos (pi/3) - sin (pi/3) + 2 cos (pi/3) sin (pi/3) = 2(1/2) - \sqrt{3}/2 + 2(1/2) \sqrt{3}/2 = 1 quad sqrt
{5pi}/3 works too
2 cos({2pi}/3) - sin({2pi}/3) + 2 cos({2pi}/3) sin({2pi}/3)
= 2(-1/2) - (\sqrt{3}/2) + 2 (-1/2)(sqrt{3}/2) = -1 - sqrt{3} quad NOPE
Similar for {4pi}/3
Final answer:
x=pi/3 or x={5pi}/3
