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Question Number 165070 by mathlove last updated on 25/Jan/22

	Answered by bobhans last updated on 25/Jan/22
	$cos x−sin x = (√(1−2sin x cos x)) ⇒((√(1−2sin x cos x))/(sin x cos x)) = 2(√2) (√(1−2t)) = 2(√2) t , [ t=sin x cos x ] ⇒1−2t = 8t^2 ⇒8t^2 +2t−1=0 ⇒(4t−1)(2t+1)=0 ⇒ { ((t=sin x cos x = (1/4); sin 2x = (1/2))),((t=sin x cos x =−(1/2) ; sin 2x=−1)) :}$
	$\cos x - \sin x = \sqrt{1 - 2 \sin x \cos x}$ $\Rightarrow \frac{\sqrt{1 - 2 \sin x \cos x}}{\sin x \cos x} = 2 \sqrt{2}$ $\sqrt{1 - 2 t} = 2 \sqrt{2} t, [t = \sin x \cos x]$ $\Rightarrow 1 - 2 t = {8 t}^{2}$ $\Rightarrow {8 t}^{2} + 2 t - 1 = 0$ $\Rightarrow (4 t - 1) (2 t + 1) = 0$ $\Rightarrow {\begin{cases} t = \sin x \cos x = \frac{1}{4}; \sin 2 x = \frac{1}{2} \\ t = \sin x \cos x = - \frac{1}{2}; \sin 2 x = - 1 \end{cases}$
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