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Question Number 18456 by Tinkutara last updated on 21/Jul/17

The complete solution of the equation

\sin 2 x - 12 (\sin x - \cos x) + 12 = 0 is

given by

(1) x = 2 n π + \frac{π}{2}, (2 n - 1) \frac{π}{4}, n \in Z

(2) x = n π + \frac{π}{2}, (2 n + 1) π, n \in Z

(3) x = 2 n π + \frac{π}{2}, (2 n + 1) π, n \in Z

(4) x = n π + \frac{π}{2}, (2 n - 1) π, n \in Z

Answered by Tinkutara last updated on 30/Jul/17

\sin 2 x = 1 - {(\sin x - \cos x)}^{2} = 1 - t^{2}

t^{2} + 12 t - 13 = 0 \Rightarrow \cos x - \sin x = - 1

\Rightarrow \cos (x + \frac{π}{4}) = - 1

\Rightarrow x = 2 n π + \frac{π}{2}, 2 n π - π