# Solving a trigonometric equation with cot

I am asked to solve the following equation: $6\sin(t)=\frac{\cos(2t)-5}{\sqrt{\tan(X)}}$ where X is the solution of $8\sin(2x)+\cos(2x)=10\cot(x)-2$.

I have already tried to replace the cotangent by equivalent expressions but I have not been able to quite solve it the way it is actually done in the textbook. Indeed, the way it is presented is by setting $z=\tan(x)$ and solving $z^3+6z^2+3z-10=0$, thus finding z to be either -5, -2 or 1. I cannot get to this last 3rd-order equation. Any hint would be much appreciated!

Rewrite $8 \sin (2 x)+\cos (2 x)=10 \cot (x)-2$ using the formulas that give $\sin$ and $\cos$ in function of $\tan$ of the half angle
$$\sin 2x =\frac{2p}{1+p^2};\;\cos 2x=\frac{1-p^2}{1+p^2};\;p=\tan x;\;\cot x = \frac{1}{p}$$ the equation becomes $$\frac{16p}{1+p^2}+\frac{1-p^2}{1+p^2}=\frac{10}{p}-2$$ Which becomes $$p^3+6 p^2+3 p-10=0$$ factoring we get $$(p-1) (p+2) (p+5)=0\rightarrow p_1=1;\;p_2=-2;\;p_3=-5$$ This means that we can substitute $\tan x=1$ but not $\;\tan x=-2;\;\tan x=-5$ in the first equation $$6\sin(t)=\frac{\cos(2t)-5}{\sqrt{\tan(x)}}$$ becomes $$6\sin(t)=\cos(2t)-5$$ and finally find the solutions for $t$.
HINT: your equation $$8\sin(2x)+\cos(2x)=10\cot(x)-2$$ is equivalent to $$2\, \left( 5\,{t}^{2}-2\,t-5 \right) \left( {t}^{2}-t-1 \right) \left( {t}^{2}+2\,t-1 \right) =0$$ with $$t=\tan(x/2)$$