Solving $\cos⁡(40+\theta) = 3\sin⁡(50+\theta)$ Question
Solve the following equation for $0≤\theta≤180$
$$
\cos⁡(40+\theta) = 3\sin⁡(50+\theta)
$$
Hint: $\cos⁡(40) = \sin(⁡50)$
Solution
$$
\cos⁡(40+\theta) = 3\sin⁡(50+\theta)
$$
$$
\cos(40)\cos\theta -\sin(40)\sin\theta = 3(\sin(50)\cos\theta +\sin\theta \cos(50))
$$
$$
\sin(50)\cos\theta-\cos(50)\sin\theta = 3(\sin(50)\cos\theta+\sin\theta \cos(50))
$$
$$
2\sin(50)\cos\theta+4\sin\theta \cos(50) = 0
$$
What do I do next?
 A: Continuing your work:
$$\sin50^\circ\cos\theta + 2\sin\theta \cos50^\circ = 0 \\ \tan\theta = -\frac{\tan50^\circ}{2} \\ \theta = -\arctan\left(\frac{\tan50^\circ}{2} \right)$$
A: Observe that $\cos(\theta)=0$ cannot give a solution. Rearranging from your last line as suggested by comments, we get $\tan(-50)=2\tan(\theta)$. Recall that $\tan(-150)=\dfrac1{\sqrt 3}$ and the trigonometric identity $\tan(3\phi)
=\dfrac{3\tan(\phi)-\tan^3(\phi)}{1-3\tan^2(\phi)}$. So $\dfrac{6\tan(\theta)-8\tan^3(\theta)}{1-12\tan^2(\theta)}=\dfrac1{\sqrt 3}\implies \dfrac1{\sqrt 3}-6\tan(\theta)-4\sqrt3\tan^2(\theta)+8\tan^3(\theta)=0$. We transform it into a monic depressed cubic in $\tan(\theta)$ as follows:
$\dfrac1{8\sqrt 3}-\dfrac34\tan(\theta)-\dfrac{\sqrt3}2\tan^2(\theta)+\tan^3(\theta)=0$. Let $x+\dfrac{\sqrt3}6=\tan(\theta),$ then the equation may be rewritten as $f(x)=x^3-x-\dfrac{\sqrt3}9=0$.
The discriminant of $f(x)$ is then $\delta^2=-4(-1)^3-27(-\dfrac{\sqrt3}9)^2=3$, and by Cardano formula, $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\sqrt{\dfrac{-3}{108}}}+\sqrt[3]{\dfrac{\sqrt3}{18}-\sqrt{\dfrac{-3}{108}}}=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}$ is a root of $f$.
Observe that $f(\dfrac{\sqrt3}3)\gt 0,f(1)\lt 0, f(2)\gt 0, $ so by intermediate value theorem, $f$ has at least two real roots. However since $f(x)\in \Bbb R[x]$, if $\alpha$ is a non real root of $f$, then so is its complex conjugate $\bar{\alpha}\ne\alpha$. Therefore $f$ has three real roots, and from this we see that $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}\in \Bbb R$.
Write $p=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}, q=\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}, $ then $\chi=p+q\implies \chi^3=(p+q)^3=p^3+q^3+3pq(p+q)$. Now, $p^3+q^3=\dfrac {\sqrt 3}9, pq=\sqrt[3]{(\dfrac{\sqrt 3}{18})^2+\dfrac 1{36}}$. Suppose that $p+q<0$, then $|p+q|^3\lt 3pq|p+q|\implies (p+q)^2=p^2+2pq+q^2\lt 3pq \implies p^2+q^2\lt pq$, but by A.M.-G.M.inequality we have $p^2+q^2\gt 2|pq|=2pq$, which gives $0\lt 2pq\lt p^2+q^2\lt pq$, a contradiction.
Hence $\chi=p+q\gt 0$ and $\tan(\theta)=\chi+\dfrac{\sqrt3}6=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6\gt 0$. Since $\arctan(\phi)\gt 0\iff \phi \gt 0, \theta=\arctan(\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6) \gt 0$, and such $\theta$ is a desired solution.
