Question

Find the number of solutions of the equation ${\sin }^2\frac{x}{2}.\sin x+1=0$.

Answer 1

For all values of $x$, $0\le {\sin }^2\left(\frac{x}{2}\right)\le 1$

and $-1\le \sin x\le 1$

${\sin }^2\left(\frac{x}{2}\right)\ne \sin x$ for all values of $x$

When $\sin x$ has a maximum or minimum value, ${\sin }^2\left(\frac{x}{2}\right)$ will have a value less than $1$

$\therefore -1<{\sin }^2\left(\frac{x}{2}\right)\sin x<1$

$\Rightarrow {\sin }^2\left(\frac{x}{2}\right)\sin x+1>0$ for all values of $x$

$\therefore {\sin }^2\left(\frac{x}{2}\right)\sin x+1=0$ has no solutions since there is no intersection.