Question

Number of solutions of the equation $\tan x+\sec x=2\cos x$ lying in the interval [0, 2$\pi$] is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

$\frac{1+\sin x}{\cos x}=2\cos x$
Or
$1+\sin \left(x\right)=2{\cos }^{2}x$
Or
$(1+\sin (x\left)\right)-2(1-{\sin }^2(x\left)\right)=0$
Or
$(1+\sin (x\left)\right)[1-2(1-\sin \left(x\right)\left)\right]=0$
Or
$(1+\sin (x\left)\right)\left(2\sin \right(x)-1)=0$
Or
$\sin \left(x\right)=-1$ and $\sin \left(x\right)=\frac{1}{2}$.
Hence
$x=\frac{3\pi }{2}$ and $x=\frac{\pi }{6},\frac{5\pi }{6}$.
Now
$\tan \left(x\right),\sec \left(x\right)$ are not defined at $x=\frac{(2n+1)\pi }{2}$ where $nϵN$.
Hence we have 2 solutions for the above expression.