Question

The set of values of $x$ satisfying the equation $1+\cos 3x=2\cos 2x$ can be
(where $n\in Z$)

• A. $n\pi \pm \frac{\pi }{3}$
• B. $n\pi \pm \frac{\pi }{6}$
• C. $n\pi \pm \frac{\pi }{4}$
• D. $2n\pi$

Answer 1

Answer: (B), (D)

$2n\pi$

$1+\cos 3x=2\cos 2x$
$\Rightarrow 1+4{\cos }^{3}x-3\cos x=4{\cos }^{2}x-2$
$\Rightarrow 4{\cos }^{3}x-4{\cos }^{2}x-3\cos x+3=0\Rightarrow 4{\cos }^{2}x(\cos x-1)-3(\cos x-1)=0\Rightarrow (4{\cos }^{2}x-3)(\cos x-1)=0$
$\Rightarrow {\cos }^{2}x=\frac{3}{4}\text{or}\cos x=1$
$x=n\pi \pm \frac{\pi }{6}\text{or}x=2n\pi ,n\in Z$