+128408 2sin x cos x + √3 cos x = 0 factor out cos x
cos x (2 sinx + √3 ) = 0
Set each factor to 0 and solve for x
cos x = 0 and this happens at x = pi/2 and 3pi/2
2sinx + √3 = 0 subtract √3 from both sides
2 sin x = -√3 divide both sides by 2
sin x = -√3 / 2 and this happens at 4pi/3 and 5pi/3
So...the correct answer is
pi/2, 4pi/3, 3pi/2, 5pi/3
