Question

The sum of all the solutions in $[0,4\pi ]$ of the equation $\tan x+\cot x+1=\cos (x+\frac{\pi }{4})$ is

• A. $3\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{7\pi }{2}$
• D. $4\pi$

Answer 1

The correct option is C $\frac{7\pi }{2}$

$\tan x+\cot x+1=\cos (x+\frac{\pi }{4})$
$\frac{1+\sin x\cos x}{\sin x\cos x}=\frac{1}{\sqrt{2}}(\cos x-\sin x)$
Squaring both sides, we get
${\sin }^{3}2x+{\sin }^{2}2x+8\sin 2x+8=0$
Put $\sin 2x=t$
$\Rightarrow t^3+t^2+8t+8=0$
$\Rightarrow (t^2+8)(t+1)=0$
$\Rightarrow t=-1,t^2=-8$
$\Rightarrow \sin 2x=-1$ $(\because {\sin }^{2}2x=-8$ (not possible) )
So $2x=\frac{3\pi }{2},\frac{7\pi }{2},\frac{11\pi }{2},\frac{15\pi }{2}$ for $x\in [0,4\pi ]$
$\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4},\frac{15\pi }{4}$for$x\in [0,4\pi ]$
But , $x=\frac{3\pi }{4},\frac{11\pi }{4}$ only satisfies the given equation.
Hence, sum of solutions $=\frac{7\pi }{2}$