We have to solve for x given sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx

sin 2x - (cos 2x)/2 -1/2 = cos x - 2cos x * tan x

=> 2 sin x cos x - (1 - 2 (sin x)^2)/2 - 1/2 = cos x - 2cos x * sin x/ cos x

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We have to solve for x given sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx

sin 2x - (cos 2x)/2 -1/2 = cos x - 2cos x * tan x

=> 2 sin x cos x - (1 - 2 (sin x)^2)/2 - 1/2 = cos x - 2cos x * sin x/ cos x

=> 2 sin x cos x - 1 + (sin x)^2 = cos x - 2 sin x

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

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

=> - cos x = 1

=> cos x = -1

=> x = arc cos -1

=> x = pi

also 2 sin x - cos x = 0

=> 2 sin x = cos x

=> 2 tan x = 1

=> x = arc tan 1/2

**Therefore x = arc tan (1/2) and x = pi**