Question

Find the number of solutions of $4cos\theta cos2\theta cos3\theta =1$ in where $\theta \in (0,\frac{\pi }{2}).$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

We have,

$4\cos \theta \cos 2\theta \cos 3\theta =1$

$\left(2\cos \theta \cos 3\theta \right)\left(2\cos 2\theta \right)=1$

$\left[\cos \right(3+\theta )+\cos (3\theta -\theta \left)\right]\left(2\cos 2\theta \right)=1$

$(\cos 4\theta +\cos 2\theta )\left(2\cos 2\theta \right)=1$

$2\cos 4\theta \cos 2\theta +2{\cos }^{2}2\theta =1$

$2\cos 4\theta \cos 2\theta +(2{\cos }^{2}2\theta -1)=0$

$2\cos 4\theta \cos 2\theta +\cos 4\theta =0$

$\cos 4\theta (2\cos 2\theta +1)=0$

$\left(i\right)\cos 4\theta =0,or\left(ii\right)2\cos 2\theta +1=0$

$\left(i\right)\cos 4\theta =0$

$\cos 4\theta =\cos (2n\pi \pm \frac{\pi }{2})$

$4\theta =(2n\pi \pm \pi /2)=\frac{(4n\pm 1)\pi }{2}$

$\theta =\frac{(4n\pm 1)\pi }{8}$

Put $n=0$

$\theta =\frac{\pi }{8}$ $\because \theta =(0,\frac{\pi }{2})$

Put $n=1$

$\theta =\frac{(4\pm 1)\pi }{8}$

$\theta =\frac{3\pi }{8}$ $\because \theta =(0,\frac{\pi }{2})$

$\left(ii\right)2\cos 2\theta +1=0$

$2\cos 2\theta =-1$

$\cos 2\theta =-\frac{1}{2}=\cos (2n\pi \pm 2\pi /3)$

$2\theta =(2n\pi \pm 2\pi /3)$

$\theta =(n\pi \pm \pi /3)$

Put $n=0$

$\theta =\frac{\pi }{3}$ $\because \theta =(0,\frac{\pi }{2})$

Hence, this equation has $3$ solution.