Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

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

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

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

$\Rightarrow 2{\sin }^{2}x+\sin x-1=0$
$\Rightarrow 2{\sin }^{2}x+2\sin x-\sin x-1=0$
$\Rightarrow 2\sin x(\sin x+2)-1(\sin x+1)=0$
$\Rightarrow (\sin x+2)(2\sin x-1)=0$
$\Rightarrow \sin x=-2$, which is not possible
or

$\sin x=\frac{1}{2}\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}$
$x\in [0,\pi ]$

$\therefore$ The number of solution $=2$.