Difficulty in solving challenging trig equation

Question

Find $\theta$ on $[0, 2\pi)$ such that

$$\cos{\theta}^{\sin{\theta}^{\cos{\theta}^{\dots}}} = 2 + 2\sec^2{\theta}\tan^2{\theta} - \sec^4{\theta} - \tan^4{\theta}$$

I'm not sure on how to tackle this problem. I've never really dealt with exponentiation of trig functions. Any help is helpful. Thank you very much

Answer 1

By inspection, $\Theta=0$ is a solution. There's little hope of solving this analytically unless you can find a closed form for the left-hand side for general $\Theta$.

Addendum: $\Theta=\pi$ is another solution because it produces the same RHS as $\Theta=0$, and the LHS $$(-1)^{0^{(-1)^0\cdots}}$$ converges to $1$ (the value of the RHS).

Answer 2

$$2 + 2\sec^2{\theta}\tan^2{\theta} - \sec^4{\theta} - \tan^4{\theta}=2+\frac{2*\sin^2\theta}{\cos^4\theta}-\frac{1}{\cos^4\theta}-\frac{\sin^4\theta}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^4\theta-2\sin^2\theta+1)}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^2\theta-1)^2}{\cos^4\theta}=\frac{2\cos^4\theta-(\cos^2\theta)^2}{\cos^4\theta}=1$$ So the whole exponent has to be 0 which is easy to tackle with(I guess you don't need help with that)

Answer 3

Working with the RHS first, we get

$$2+2\sec^2\theta\tan^2\theta-\sec^4\theta-\tan^4\theta\\=2-(\sec^2\theta-\tan^2\theta)^2=2-1=1$$

Now we have an easier relation to move forward with:

$$\cos\theta^{\sin\theta^{\cos\theta^{\dots}}}=1$$

But $\cos\theta=1$ exactly when $\theta=0+2k\pi$, and $\theta^x=0$ only when $\theta=0$. For any other solution to exist, we must have some $\theta,x$ such that $\theta^x=2k\pi$, and since $\sin \theta\in[-1,1]$ we end up with $[0,\pi)^{[0,1)}\subset [0,\pi)$ and $[\pi,2\pi)^{[-1,1]}\subset [0,2\pi)$. In the first case, $0$ is the only possible solution and in the second case, there is no possible solution.

Therefore $\theta =0$ is the only possible solution using just $\cos\theta=1$. Follow the answer by @MPW for the additional solution $\theta=\pi$ to give $(\cos\theta)^0=1$.