MCQ
If $\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x}),$ then $\sin2\text{x}=$
- A$\pm\frac34$
- B$\pm\frac43$
- C$\pm\frac13$
- DNone of these
Solution:
$\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x})$
As we know that
$\sin\text{x}=-\cos\Big(\frac\pi2+\text{x}\Big)$$\Rightarrow-\cos\Big(\frac\pi2+\pi\cos\text{x}\Big)=\cos(\pi\sin\text{x})$
$\Rightarrow\frac{-\pi}{2}-\pi\cos\text{x}=\pi\sin\text{x}$
$\Rightarrow\pi\sin\text{x}-\pi\cos\text{x}=\frac12$
$\Rightarrow\sin\text{x}-\cos\text{x}=\frac12$
Squaring both sides we get,
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\frac14$
$\Rightarrow1+\sin2\text{x}=\frac{1}{4}$
$\Rightarrow\sin2\text{x}=\frac13$
$\therefore\sin2\text{x}=\pm\frac13$
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Domain $= [1, \infty),$ Range $= [0, \infty)$
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