Trigonometry! #57

Geometry Level 2

The number of roots of the equation $3\sin^{2}x=8\cos x$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is

This problem is part of the set Trigonometry .

1 solution

Omkar Kulkarni
Feb 20, 2015

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

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

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

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

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

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

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

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

$\boxed{x=-\cos^{-1}\left(\frac{1}{3}\right),\cos^{-1}\left(\frac{1}{3}\right)}$

Your second line is missing a closing bracket and it should be noted that $|cosx|<1$ , hence why $cosx=-3$ is an extraneous solution.

Marvin Chong - 1 year, 4 months ago