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