Solution of $\theta$ when $\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$ I came across this trigonometry problem. If, $$\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$$
What is the value of $\theta$
I got the solution that $\theta$ will be $\frac{\pi}{3}$ by expanding the equation and turning into,
$$\sin^4(\theta)+\sqrt3\sin^3(\theta)+\frac{3}{4}\sin^2(\theta)-\sqrt3\sin(\theta)-\frac{3}{4}=0$$
And then solving $\sin(\theta)$. But this method seems too complicated. Is there any easier and better solution?

$$\sqrt{3}\,t^4 + 8\,t^3 - \sqrt{3} =\left(t-\frac{1}{\sqrt{3}}\right) \left(\sqrt{3} t^3+9 t^2+3 \sqrt{3} t+3\right).$$
Using the hyperbolic method for the cubic gives for the real root
$$t=-\sqrt{3}-2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)$$ from which
$$\theta=2\pi -2 \tan ^{-1}\left(\sqrt{3}+2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)\right).$$
