Question

Number of solutions of $\tan x+\sec x=2\cos x$ in intarval $[0,2\pi ]$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$\tan x+\sec x=2\cos x$

$\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$

$\Rightarrow \frac{(1+\sin x)}{\cos x}=2\cos x$

$\Rightarrow 1+\sin x=2{\cos }^{2}x$

$\Rightarrow (1+\sin x)=2(1-{\sin }^{2}x)$

$\Rightarrow (1+\sin x)-2(1-\sin x)(1+\sin x)=0$

$\Rightarrow (1+\sin x)(1-2+2\sin x)=0$

$\Rightarrow (1+\sin x)\left(2\sin x-1\right)=0$

$\Rightarrow \sin x=-1,\sin x=\frac{1}{2}$

$\therefore$ Between $[0,2\pi ],$ there can be three solutions $\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}.$

But $x=\frac{3\pi }{2}$ is not satisfying the original equation.So,

Hence, the answer is $\frac{\pi }{6},\frac{5\pi }{6}$