$e^{\cot^2\theta}+\sin^2\theta-2\cos^22\theta+4=4\sin\theta$ in $[0,10\pi]$

Question

Number of solution of the equation $e^{\cot^2\theta}+\sin^2\theta-2\cos^22\theta+4=4\sin\theta$ in $[0,10\pi]$ is $(A)2\hspace{1cm}(B)3\hspace{1cm}(C)4\hspace{1cm}(D)5$

In this question,$e^{\cot^2\theta}$ varies from $1$ to infinity,whereas $RHS$ varies from $-4$ to $4$.How can i find the number of solutions without using graphing calculator.Please help me.

Answer 1

It's easily checkable that $n\pi$ is not a solution. LHS and RHS have periods of $\pi,2\pi$ respectively so we only need to worry about the interval $(0,2\pi)$. If $\xi$ is a solution of the equation in $(0,2\pi)$, $\xi+2n\pi$ is also a solution in $(2n\pi,2(n+1)\pi)$. So at least one solution in $(0,2\pi)$ guarantees at least $5$ solutions in $[0,10\pi]$ and no solution in $(0,2\pi)$ implies $0$ solutions in $[0,10\pi]$. Therefore using the information from the options, the correct choice is $(D)$.

Answer 2

We rewrite the equation as $e^{\cot^2\theta}=2\cos^22\theta-\sin^2\theta+4\sin\theta-4$. The equation has a period of 2π. L.H.S.≥1, then R.H.S.≥1, i.e. $2\cos^22\theta-\sin^2\theta+4\sin\theta-4≥1$, then we have $\cos4\theta≥(sin\theta-2)^2$. Since $\cos4\theta$∈[-1,1] and $(sin\theta-2)^2$∈[1,9] for θ∈[0,2π], θ must be π/2. Therefore we can see that the equation has solutions {π/2,5π/2,9π/2,13π/2,17π/2} for θ∈[0,10π].