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solve the equation
2 sin² x = 3 cos x , for 0 <= x <= 5π
2 sin² x = 2(1-cos² x)
so equation becomes
2 - 2cos² x = 3 cos x
2cos² x + 3cos x - 2 = 0
(2cos x - 1) (cos x + 2) = 0
cos x = 1/2 or cos x = -2 (reject)
so x = π/3, 5π/3, 7π/3, 11π/3, 13π/3
