Solve cos2 x greaterthanorequalto sin 2x 0 lessthanorequalto

Solve: cos^2 x greaterthanorequalto sin 2x, 0 lessthanorequalto x lessthanorequalto 2 pi

Solution

cos^2 x >= sin 2x

sin 2x = 2 sin x cos x

so we can write

cos^2 x >= 2 sin x cos x

cos ^2 x - 2 sin x cos x >= 0

cos x ( cos x - 2 sin x ) >= 0

cos x >= 0

cos x - 2 sin x >= 0

cot x >= 2

x = 2pi*n +pi = pi

2pi*n - pi < x <= 2 pi*n - 2tan^-1 ( 2+ sqrt 5 )

1/4 ( 4pi*n - pi ) <= x <= 2pi*n - 2tan^-1 ( 2+ sqrt 5 )

1/2 ( 4pi*n - pi ) <= x < 2pi*n + pi