Question

The number of roots of the equation $si{n}^{-1}x-co{s}^{-1}x=si{n}^{-1}\left(5x-3\right)$ is/are

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

Answer: (B)

$si{n}^{-1}x-co{s}^{-1}x=si{n}^{-1}\left(5x-3\right)$
$\Rightarrow \frac{\pi }{2}-co{s}^{-1}x-co{s}^{-1}x=\frac{\pi }{2}-co{s}^{-1}\left(5x-3\right)$
$\Rightarrow 2co{s}^{-1}x=co{s}^{-1}\left(5x-3\right).Alsox\in \left[-1,1\right]$ … $\left(1\right)$
$\Rightarrow co{s}^{-1}\left(2x^2-1\right)=co{s}^{-1}\left(5x-3\right)$ and $\left(5x-3\right)\in \left[-1,1\right],$ i.e., $-1\le 5x-3\le 1$
$\Rightarrow 2x^2-1=5x-3,hence,x\in \left[\frac{2}{5},\frac{4}{5}\right]$
$\Rightarrow 2x^2-5x+2=0\Rightarrow x=2$ or $\frac{1}{2}$
but, $x=2$ does not satisfy the equation $\left(1\right)$
Hence, the given equation has only one root