Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval

tan x + sec x = 2 cos x in  [0, 2π]

i.e $\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x$ where $x \notin(2 x+1) \frac{\pi}{2}$        ( $\because \cos x$ is not defined)

i.e sin x + 1 = 2 cos2x

⇒ sin x + 1 = 2(1 − sin2x)               (∵ sin2x + cos2x = 1)

i.e 2 sin2x + sin x − 1 = 0

i.e 2 sin2x + 2 sin x – sin x – 1 = 0

i.e 2 sin x (sin x + 1) −1 (sin x + 1) = 0

i.e (2 sin x − 1) (sin x + 1) = 0

i.e $\sin x=\frac{1}{2}$ or $\sin x=-1$

i.e for $\sin x=\frac{1}{2}$

we have $x$ is I or II Quadrant

i.e $x=\frac{\pi}{6}$ or $\pi-\frac{\pi}{6}=\frac{\sqrt{\pi}}{6}$ and for $\sin x=-1$

$x=\frac{3 \pi}{2}$ which is not possible $\quad \because x \notin(2 x+1) \frac{\pi}{2}$

Hence, tan x + sec x = 2 cos x has 2 columns in [0, 2π">ππ]

Hence, the correct answer is option C.