I start simplifying the L.H.S of the above equation as follows: $$y=\log_{\sqrt{3}}\tan(\theta)\sqrt{\log_{\tan(\theta)}(3)+\log_{\sqrt{3}}3\sqrt{3}}=\log_{\sqrt{3}}\tan(\theta)\sqrt{\frac{\log_{\sqrt{3}}(3)}{\log_{\sqrt{3}}\tan(\theta)}+\log_{\sqrt{3}}(\sqrt{3})^3}$$ $$y=\log_{\sqrt{3}}\tan(\theta)\sqrt{\frac{2}{\log_{\sqrt{3}}\tan(\theta)}+3}$$ Since $y=-1$, with some hindsight I figured out two possible solutions: $$\theta=\frac{\pi}{6}\text{ or } \frac{7\pi}{6}$$ However, without using Wolfram I could not establish the uniqueness of these solutions. I have no way of determining, intuitively or otherwise if this equation has more solutions. Since the equation is not polynomial, how do I determine the uniqueness of my solutions by hand? To the best of my knowledge, equations like these do not have the notion of a degree to establish the maximum possible number of solutions.
Solutions of $\log_{\sqrt{3}}\tan(\theta)\sqrt{\log_{\tan(\theta)}(3)+\log_{\sqrt{3}}3\sqrt{3}}=-1$ for $\theta\in [0,2\pi]$.
algebra-precalculussolution-verificationtrigonometry
Best Answer
Let $u = \log_{\sqrt{3}} (\tan \theta)$. Then your equation is as follows:
$$u\sqrt{\frac{2}{u} + 3} = -1 \Rightarrow u^2 (2/u + 3) = 1 \Rightarrow 3u^2+2u-1 = 0 \Rightarrow (3u-1)(u+1) = 0 \Rightarrow u = -1, \frac{1}{3}$$
Now since we have squared both sides, substitute both solutions into the original inequality. $u = -1$ is valid, whereas $u = \frac{1}{3}$ is not.
Then you can raise both sides to the power of $\sqrt{3}$ to get the value of $\tan \theta$. Can you continue?