Question

Find the number of roots of the equation $3{\sin }^{2}x=8\cos x$ in $[\frac{-\pi }{2},\frac{\pi }{2}]$

Answer 1

$3{\sin }^{2}x=8\cos x$

$3(1-{\cos }^{2}x)=8\cos x$

$3-3{\cos }^{2}x-8\cos x=0$

$3{\cos }^{2}x+8\cos x-3=0$

$3{\cos }^{2}x+9\cos x-\cos x-3=0$

$3\cos x(\cos x+3)-1(\cos x+3)=0$

$(\cos x+3)(3\cos x-1)=0$

$\cos x\ne -3$ since range of cosine function is $[-1,1]$

$\therefore \cos x=\frac{1}{3}$

$\therefore x=\pm {\cos }^{-1}\frac{1}{3}$

Thus there are $2$ solutions.