$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$ $\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$

$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\frac{\sin x\cos x}{\sqrt{\sin^2x+\sin^2\alpha}})-\sin x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$
I am stuck here and could not find the minimum and maximum values of $f(x),$The answer given is $-\sqrt{1+\sin^2\alpha}\leq f(x)\leq\sqrt{1+\sin^2\alpha}$.

You're derivative need work. $$f'(x)=-\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})\left(\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alpha}}\right).$$
Now determine when this is zero, which is precisely when $$\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})=0,$$ or $$\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alpha}}=0.$$
