Question

The number of solutions of the equation $\cos 6x+{\tan }^{2}x+\cos 6x\cdot {\tan }^{2}x=1$ in the interval $[0,2\pi ]$ is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

7

$cos6x+ta{n}^{2}x+cos6x.ta{n}^{2}x=1$

$\Rightarrow cos6x(1+ta{n}^{2}x)=(1-ta{n}^{2}x)\Rightarrow cos6x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}$

$\Rightarrow cos6x=cos2x$ $[\because \sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$, $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}]$

Threrefore general solution is,

$6x=2n\pi \pm 2x$, where $n$ is any integer

$x=\frac{n\pi }{2}$ and $x=\frac{n\pi }{4}$

Hence solutions in the given interval are,

$x=0,\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\pi ,\frac{3\pi }{4},2\pi$ , total $7$ solutions.