Question

Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

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 ∉(2x + 1) $\frac{\mathrm{\pi }}{2}$ (∵ 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 = $\frac{\mathrm{\pi }}{6}\mathrm{or}\mathrm{\pi }-\frac{\mathrm{\pi }}{6}=\frac{\sqrt{\mathrm{\pi }}}{6}$ and for sin x = −1
x = $\frac{3\mathrm{\pi }}{2}$ which is not possible $\because x\notin \left(2x+1\right)\frac{\mathrm{\pi }}{2}$
Hence, tan x + sec x = 2 cos x has 2 columns in [0, 2$\mathrm{\pi }$]

Hence, the correct answer is option C.