We have to solve (1 + cos 2x) / 2 = sin (-2x)

(1 + cos 2x) / 2 = sin (-2x)

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

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

=> -1...

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We have to solve (1 + cos 2x) / 2 = sin (-2x)

(1 + cos 2x) / 2 = sin (-2x)

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

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

=> -1 - (cos x)^2 + (sin x)^2 = 4 sin x cos x

=> -(cos x)^2 - (sin x)^2 - (cos x)^2 + (sin x)^2 =4 sin x cos x

=> - 2(cos x)^2 = 4 sin x cos x

=> -2 cos x = 4 sin x

=> sin x / cos x = -2/4

=> tan x = -1/2

x = arc tan (-1/2)

**Therefore x = arc tan ( -1/2) + n*pi**