Question

Number of solutions of the equation $tanx + secx = 2cosx$ lying in the interval $[0,2\pi]$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

Answer 2

Given equation is $tanx + secx = 2cosx$ $\Rightarrow \frac{sin\,x}{cos\,x}+\frac{1}{cos\,x}=2cosx$ $\Rightarrow 1+sinx=2cos^{2}x$ $\Rightarrow 1+sinx=2\left(1-sin^{2}x\right)$ $\Rightarrow 2sin^{2}x+2sinx-sinx-1=0$ $\Rightarrow 2sinx\left(sinx + 1\right) - 1\left(sinx +1\right) = 0$ $\Rightarrow \left(2sinx - 1\right)\left(sinx +1\right) = 0$ $\Rightarrow$ either $sinx=\frac{1}{2}$ or $sinx=-1$ $\Rightarrow$ either $x=\frac{\pi}{6}, \frac{5\pi}{6} \in\left[0, \pi\right]$ or $x=\frac{3\pi}{2}$ But $x=\frac{3\pi}{2}$ can not be possible. $\therefore$ Number of solutions are $2$.