Mathematics - Functions Question with Solution | TestHub

$|\sin x-1|\ne 0$

$$\begin{array}{rl}& \sin x\ne 1\\ & |\sin x-1|\le 1\\ & -1\le \sin x-1\le 1\\ & 0\le \sin x\le 2\end{array}$$

From (i) and (ii), $\sin x\in [0,1)$

$$\begin{array}{rl}& \sin x\in [0,1)\\ & \sin x=0\to x=n\pi \end{array}$$

$$\sin x=1\to x=n\pi +(-1{)}^n\frac{\pi }{2}$$

⇒ Domain of $f(x)$ is

$$\begin{array}{ll}x=n\pi (n\text{ any integer })& \\ \sin x\le 1& x\in \left[0,\frac{\pi }{2}\right)\end{array}$$

General solution is

$$x=n\pi +(-1{)}^n\alpha$$

where, $\alpha \in \left[0,\frac{\pi }{2}\right)$.