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

The correct option is D 3

$tanx+secx=2cosx$
$\Rightarrow sinx+1=2co{s}^{2}x\Rightarrow sinx+1=2-2si{n}^{2}x$
$\Rightarrow 2si{n}^{2}x+sinx-1=0$
$\Rightarrow (2sinx-1)(sinx+1)=0$
but $sinx=-1\Rightarrow x=\frac{3\pi }{2}$ ...........(1)
$sinx=\frac{1}{2}=sin\frac{\pi }{6}$
therefore general solution is,
$x=n\pi +(-1{)}^n.\frac{\pi }{6}$
$x=......,\frac{\pi }{6},\frac{5\pi }{6},......$ ..........(2)
Therefore, number of solutions in the given interval are $3$.
Hence, option 'D' is correct.