Question

The trigonometric equation $si{n}^{-1}x=2si{n}^{-1}$ a has a solution for:

• A. all real values of a
• B. $|a|<\frac{1}{2}$
• C. $|a|\le \frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}<|a|<\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $|a|\le \frac{1}{\sqrt{2}}$

The equation $si{n}^{-1}x=2si{n}^{-1}$ a has solution if $-\frac{\pi }{2}\le 2si{n}^{-1}a\le \frac{\pi }{2}$
i.e., $-\frac{\pi }{4}\le si{n}^{-1}a\le \frac{\pi }{4}$ i.e., $\left(\frac{1}{\sqrt{2}}\right)\le a\le \frac{1}{\sqrt{2}}$ i.e., $|a|\le \frac{1}{\sqrt{2}}$