Question

If $\frac{\cos x}{(1+\sin x)}=\left|\tan x\right|$, then find the number of solutions in $[0,2\pi ]$.

Answer 1

Given:

$\frac{\cos x}{(1+\sin x)}=\left|\tan x\right|$

There will be two cases to check the solution of $\left|\tan x\right|$ in $[0,2\pi ]$

Case 1: $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{x}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{0}$

$\frac{\cos x}{(1+\sin x)}=\frac{\sin x}{\cos x}$

$\Rightarrow$ ${\cos }^{2}x={\sin }^{2}x+\sin x$

$\Rightarrow$ $1-{\sin }^{2}x={\sin }^{2}x+\sin x$

$\Rightarrow$$2{\sin }^{2}x+\sin x–1=0$

$\Rightarrow$ $\sin x=-1,\frac{1}{2}$

$\Rightarrow$ $\sin x=-1$ (not possible)

$\Rightarrow$ $\sin x=\frac{1}{2}$

$\Rightarrow$ $x=\frac{\pi }{6}$

Case 2: $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{x}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}$

$\frac{cosx}{(1+\sin x)}=-\frac{\sin x}{\cos x}$

$\Rightarrow$ ${\cos }^{2}x=-{\sin }^{2}x–\sin x$

$\Rightarrow$$1-{\sin }^{2}x=-{\sin }^{2}x–\sin x$

$\Rightarrow$ $sinx=-1$ (Not possible)

$\therefore$Only one solution exists $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{6}}$