Question

The value(s) of $\theta$, which satisfy the equation : $2{\cos }^{3}3\theta +3\cos 3\theta +4=3{\sin }^{2}3\theta$ is/are

• A. $\frac{2n\pi }{3}+\frac{2\pi }{9},n\in I$
• B. $\frac{2n\pi }{3}-\frac{2\pi }{9},n\in I$
• C. $\frac{2n\pi }{5}+\frac{2\pi }{5},n\in I$
• D. $\frac{2n\pi }{5}-\frac{2\pi }{5},n\in I$

Answer 1

Answer: (A), (B)

$\frac{2n\pi }{3}-\frac{2\pi }{9},n\in I$
$\frac{2n\pi }{3}+\frac{2\pi }{9},n\in I$

$2{\cos }^{3}3\theta +3\cos 3\theta +4=3{\sin }^{2}3\theta$

$\Rightarrow (\cos 3\theta +\frac{1}{2})(2{\cos }^{2}3\theta +2\cos 3\theta +2)=0$

$\Rightarrow (\cos 3\theta +\frac{1}{2})=0$ or $(2{\cos }^{2}3\theta +2\cos 3\theta +2)=0$

If $2{\cos }^{2}3\theta +2\cos 3\theta +2=0$, then we get the value of $\cos 3\theta$ in complex form

which is not possible, as $-1\le \cos 3\theta \le 1$

$\Rightarrow \cos 3\theta =-\frac{1}{2}$

$\Rightarrow 3\theta =2n\pi \pm \frac{2\pi }{3}$

$\Rightarrow \theta =\frac{2n\pi }{3}\pm \frac{2\pi }{9}$