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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)) :}

$$\:\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:=\:\sqrt{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}} \\ $$$$\:\Rightarrow\frac{\sqrt{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}}{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−\mathrm{2t}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{t}\:,\:\left[\:\mathrm{t}=\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\right] \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2t}\:=\:\mathrm{8t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{8t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{4t}−\mathrm{1}\right)\left(\mathrm{2t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{t}=\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}};\:\mathrm{sin}\:\mathrm{2x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{t}=\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:;\:\mathrm{sin}\:\mathrm{2x}=−\mathrm{1}}\end{cases} \\ $$$$ \\ $$

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