Question

General solution of $x$ for which ${\cos }^{2}x=\frac{\sin x\tan x}{6}$ is :

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

Answer 1

The correct option is B $2n\pi \pm \frac{\pi }{3},n\in Z$

${\cos }^{2}x=\frac{\sin x\tan x}{6}\Rightarrow 6{\cos }^{3}x=1-{\cos }^{2}x\Rightarrow 6{\cos }^{3}x+{\cos }^{2}x-1=0$
Let $\cos x=t,$ where $t\in [-1,1]$
$\Rightarrow 6t^3+t^2-1=0...\left(1\right)$
From the given options, we have
$\cos \frac{\pi }{3}=\frac{1}{2},\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

We observe that only $t=\frac{1}{2}$ is satisfied the equation $\left(1\right).$
$(\because \frac{6}{8}+\frac{1}{4}-1=0)$

Now, after dividing the equation $\left(1\right)$ form $(2t-1),$ we get
$6t^3+t^2-1=(2t-1)(3t^2+2t+1)$
So, for remaining roots $3t^2+2t+1=0$
But $\Delta <0$
$\Rightarrow$ No real roots of the quadratic equation.

So, only possible real root of the cubic equation is $\cos x=\frac{1}{2}$
$\therefore x=2n\pi \pm \frac{\pi }{3},n\in Z$