# The number of solutions of the equation

Question:

The number of solutions of the equation

$|\cot x|=\cot x+\frac{1}{\sin x}$ in the interval $[0,2 \pi]$ is

Solution:

If $\cot x>0 \Rightarrow \frac{1}{\sin x}=0$ (Not possible)

If $\cot x<0 \Rightarrow 2 \cot x+\frac{1}{\sin x}=0$

$\Rightarrow 2 \cos x=-1$

$\Rightarrow x=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$ (reject)
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}