Question

The 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: (D)

Given, $tanx+secx=2cosx$
$\frac{\left(1+sinx\right)}{cosx}=2cosx$
$1+sinx=2co{s}^{2}x$
$1+sinx=2\left(1-si{n}^{2}x\right)$
$1+sinx=2-2si{n}^{2}x$
$2si{n}^{2}x+sinx-1=0$
$2si{n}^{2}x+2sinx-sinx-1=0$
$2sinx\left(sinx+1\right)-1\left(sinx+1\right)=0$
$\left(sinx+1\right)\left(2sinx-1\right)=0<br>sinx=-1,12$
$sinx=sin270^{\circ },sin30^{\circ },sin150^{\circ }$
$\Rightarrow x=\pi /6,5\pi /6,3\pi /2$
Hence there are three solutions of given equation in $\left[0,2\pi \right]$