Two Lines Enclose A Cot

Either $\cot(x) = \cot(x) + \sin(x)$ or $\cot(x) = -\cot(x) - \sin(x)$ .

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First Case - $\cot(x) = \cot(x) + \sin(x)$

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Simplifying the equation, we have $\sin(x) = 0$ . But since $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$ where $\sin(x) \neq 0$ , this proves that there are no solutions.

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Second Case - $\cot(x) = -\left(\cot(x) + \sin(x)\right)$

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Solving the equation, $\begin{array}{rlcccl} \dfrac{\cos(x)}{\sin(x)} &= -\dfrac{\cos(x)}{\sin(x)} - \sin(x) & & & & {\color{#3D99F6}\cot(x) = \dfrac{\cos(x)}{\sin(x)}}\\ \cos(x) &= -\cos(x) - \sin^2(x)\\ 0 &= -2\cos(x) - \sin^2(x)\\ 0 &= -2\cos(x) - \left(1 - \cos^2(x)\right) & & & & {\color{#3D99F6}\text{Since }\sin^2(x) + \cos^2(x) = 1}\\ 0 &= \cos^2(x) - 2\cos(x) - 1\\ 1 &= \cos^2(x) - 2\cos(x)\\ 2 &= \cos^2(x) - 2\cos(x) + 1 & & & & {\color{#3D99F6}\text{Completing the squares}}\\ 2 &= \left(\cos(x) - 1\right)^2\\ \pm\sqrt{2} &= \cos(x) - 1\\ x &= \arccos\left(1 \pm \sqrt{2}\right) \end{array}$ Since $1 + \sqrt{2} > 1$ , this gives imaginary solution. Therefore, $x = \arccos\left(1 - \sqrt{2}\right)$ , which is within $(0,\pi)$ .

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Final Result

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Thus, there is $\boxed{\text{one solution}}$ .

Let $f(x) = |\cot{x}|-\cot{x}-\sin{x}$

$f(x) = \left( \begin{matrix} \text{if } 0<x\le\frac{\pi}{2} & f(x) =\cancel{\cot{x}}-\cancel{\cot{x}}-\sin{x} \\ \text{if } \frac{\pi}{2}< x < \pi & f(x) = -2\cot{x}-\sin{x} \end{matrix} \right)$

• $0<x\le\dfrac{\pi}{2}$

$\sin{x} = 0 \implies x = n\pi \quad \small\text{where n is an integer} . \\ x = \phi$

• $\dfrac{\pi}{2}< x < \pi$

$\begin{aligned} -2\cot{x} & = \sin{x} \\ -2\cos{x} & = 1-cos^{2}{x} \quad \small\color{#3D99F6}{\text{Cosx should be -ve }} \\ \cos{x} & = 1+\sqrt2,1-\sqrt2 \\ x &= \cos^{-1}(-0.414) \end{aligned}$

$\text{Only 1 solution exist}$