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?
 A: One good method is to use the Weierstrass substitution
$$ \sin(\theta) = \frac{2t}{1+t^2},\quad 
\cos(\theta) = \frac{1-t^2}{1+t^2},
\quad \tan\left(\frac{\theta}2\right) = t. $$
Your original equation to solve is
$$ \tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}. $$
Substitute using $\,t\,$ to get
$$ \frac{2t}{1-t^2} - \frac{2t}{1+t^2} = \frac{\sqrt3}{2}. $$
Solving this leads to a quartic equation
$$ \sqrt{3}\,t^4 + 8\,t^3 - \sqrt{3} = 0 $$
which has two conjugate complex  roots, and one of the real
roots is $\,\frac1{\sqrt{3}}\,$ which yields $\,\theta=\frac{\pi}3.$
The other real root is not so nice. It is $\,t = -4.628884\dots\,$
which yields $\,\theta = 3.567122\dots.$

However, if you think you are lucky, then you can find
one real solution by inspection. You can use the known facts that
$\,\sin(\pi/3) = \sqrt{3}/2\,$ and $\,\tan(\pi/3) = \sqrt{3}\,$ to
get one obvious solution $\,\theta=\pi/3\,$ although his won't get
you any closer to the other three solutions.
A: $$\sqrt{3}\,t^4 + 8\,t^3 - \sqrt{3} =\left(t-\frac{1}{\sqrt{3}}\right) \left(\sqrt{3} t^3+9 t^2+3 \sqrt{3} t+3\right).$$
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
$$\theta=2\pi -2 \tan ^{-1}\left(\sqrt{3}+2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)\right).$$
A: There are two real solutions between $0$ and $2\pi$: namely, $\pi/3 \approx 1.047$ and
$$
\arctan \left( (1/2)\, \left( \sqrt [3]{7+4\,\sqrt {3}}- \left( 7+4\,
\sqrt {3} \right) ^{2/3} \right)  \left( \sqrt {3}-2 \right)  \right) 
+\pi \approx 3.56712 .
$$
(There are two complex solutions as well.)
