The number of roots for this equation

Question

To find the number of roots of $2\cos^{2} \theta=1+3^{\sin \theta}+3^{\csc \theta}$ in the interval [0,$2\pi$]

I was only able to simplify it to

$\cos 2\theta=3^{\sin \theta}+3^{\csc \theta}$

I'm uncertain on how to proceed after this .

Thanks in advance guys.

Answer 1

Partial: You have that $$3^{\sin(\theta)}+3^{\csc(\theta)}=\cos(2\theta)\leq 1$$ So:

$$3^{\sin(\theta)}\leq 1 \Rightarrow \sin(\theta)\leq 0$$ But $\sin(\theta)\neq 0$(otherwise cosecant is not defined), so $\sin(\theta)<0$

So a first restriction is $\theta \in (\pi,2\pi)$ Notice now that the LHS function, and the RHS function are symmetrical respect to $\theta=\frac 32 \pi$. So you can restrict even more to $(\pi,\frac 32 \pi)$(I verified and $\frac 32 \pi$ is not a solution). Moreover $cos(2\theta)$ has to be $>0$(it's sum of two exponentials in base 3). So our interval reduces even more to $(\pi,\frac 54 \pi)$.