Find number of solutions of the given equation in the specified interval

Question

I have absolutely no idea how to approach the following question -

Q: Find the number of solutions of the following equation: $$1+e^{\cot^2x} = \sqrt{2| \sin x | -1} + \frac{1-\cos2x}{1 + \sin^4x}$$ for $x\in(0,5\pi)$.

Answer 1

Let us start by finding the ranges of the functions on both sides. To that end we proceed.

RHS: $$|\sin x| \in [0,1]$$ $$\sqrt{2|\sin x|-1} \text{ is defined only when} |\sin x| \in \left[\frac12, 1\right]$$ Thus, range of $\sqrt{2|\sin x|-1}$ in this range of $|\sin x|$ is $[0,1]$.

Now, if $$|\sin x| \in \left[\frac12, 1\right]$$ $$\cos 2x = 1 - 2\sin^2 x \in \left[-1,\frac12\right]$$ $$\text{ and } \sin^4 x \in \left[\frac1{16},1\right]$$ $$\implies \frac{1-\cos 2x} {1+\sin^4 x} \in \left[\frac8{17},1\right]$$

Thus, $$\sqrt{2|\sin x|-1} + \frac{1-\cos 2x}{1+\sin^4 x} \in \left[\frac8{17},2\right] \tag 1$$

LHS: For $$|\sin x| \in \left[\frac12, 1\right] \implies \cot^2 x \in \left[0,3\right]$$ $$\implies 1+e^{\cot^2 x} \in \left[2,1+e^3\right] \tag 2$$

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This gives us $$(1) \land (2) \equiv \{2\}$$ $$\implies \cot^2 x = 0 $$ $$\boxed{x = \frac{(2n+1)\pi}2\,,\, n=\{0,1,2,3,4\}}$$ confirming the Doctor's computations.

Answer 2

i have found $$\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{\pi}{2};\frac{9\pi}{2}$$