Question

The number of solutons of the equation tan x +sec x = 2cos x lying in the interval $[0,2\pi ]$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C

2

The given equation is
tan x +sec x =2 cos x
or $\frac{sinx}{cosx}+\frac{1}{cosx}=2cosx$
or $sinx+1=2co{s}^{2}x=2-2si{n}^{2}x$
or $2si{n}^{2}x+sinx-1=0$
or (2 sinx-1)(sin x+1)=0
or $sinx=\frac{1}{2},-1$
$\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}\in [0,2\pi ]$
But for $x=\frac{3\pi }{2}$, tan x and sec x are not defined.
Therefore, there are only two solutoins.