The number of solutions of the equation
$|\cot x| = \cot x + {1 \over {\sin x}}$ in the interval [ 0, 2$\pi$ ] is
Answer (integer)
1
Solution
Case I : When cot x > 0, $$x \in \left[ {0,{\pi \over 2}} \right] \cup \left[ {\pi ,{{3\pi } \over 2}} \right]$$<br><br>$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow$ not possible<br><br>Case II : When cot x < 0, $$x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$$<br><br>$- \cot x = \cot x + {1 \over {\sin x}}$<br><br>$\Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$<br><br>$\Rightarrow \cos x = {{ - 1} \over 2}$<br><br>$\Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$(Rejected)<br><br>One solution.
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Equations
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