Question

Which of the following sets can be the subset of the general solution of the equation $1+\cos 3x=2\cos 2x$ ?

• A. $\{n\pi +\frac{\pi }{3},n\in Z\}$
• B. $\{n\pi +\frac{\pi }{6},n\in Z\}$
• C. $\{n\pi -\frac{\pi }{6},n\in Z\}$
• D. $\{2n\pi ,n\in Z\}$

Answer 1

Answer: (B), (C), (D)

$\{2n\pi ,n\in Z\}$

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