Solve for x equation sinx*cosx=2sin^2x? Solution We\'ll subtract 2 [sin^(2)] x both sides: sin x* cos x - 2 [sin^(2)] x = 0 We\'ll factorize by sin x: sin x(cos x - 2sin x) = 0 We\'ll cancel each factor: sin x = 0 x = [(-1)^(k)] arcsin 0 + k [pi] x = k [pi] cos x - 2sin x = 0 We\'ll divide by cos x both sides: 1 - 2 tan x = 0 -2 tan x = -1 tan x = 1/2 x = arctan (1/2) + k [pi] x = 0.147 [pi] + k [pi] The solutions of the equation belong to the reunion of sets {k [pi] / k [in] Z} U {0.147 [pi] + k [pi] / k [in] Z}..