The trigonometric equation sin ^{-1} x = 2 si | vedclass

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(D) We know that the range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\sin ^{-1} x = 2 \sin ^{-1} 2a$,the value of $2 \sin ^{-1} 2a

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$|a| \leq \frac{1}{2 \sqrt{2}}$

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(D) We know that the range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\sin ^{-1} x = 2 \sin ^{-1} 2a$,the value of $2 \sin ^{-1} 2a$ must lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.$\Rightarrow -\frac{\pi}{2} \leq 2 \sin ^{-1} 2a \leq \frac{\pi}{2}$$\Rightarrow -\frac{\pi}{4} \leq \sin ^{-1} 2a \leq \frac{\pi}{4}$Taking sine on all sides,we get:$\sin(-\frac{\pi}{4}) \leq 2a \leq \sin(\frac{\pi}{4})$$\Rightarrow -\frac{1}{\sqrt{2}} \leq 2a \leq \frac{1}{\sqrt{2}}$Dividing by $2$,we get:$-\frac{1}{2\sqrt{2}} \leq a \leq \frac{1}{2\sqrt{2}}$This is equivalent to $|a| \leq \frac{1}{2\sqrt{2}}$.

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} 2a$ has a real solution,if

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