I'm trying to find all the real solutions of the following trigonometric equation :
$$ \frac{\sin 3x}{\sin 2x} = A $$
I can see that $x$ is some integer function of $\pi$ , but I cannot find the exact expression for it using normal trigonometric identities due to $A$ being present.
Any help would be greatly appreciated!
With Double Angle Identity & Triple Angle Identity , we have :
$$\frac{\sin 3x}{\sin 2x}=\frac{-4\sin^3x+3\sin x}{2\sin x\cos x}=\frac{-4\sin^2 x+3}{2\cos x}=A$$
So , we get the Equation :
$$4\cos^2 x-2A\cos x-1=0$$
It is Quadratic in $\cos x$.
Then , the Quadratic Solutions are :
$$\cos x=\frac{A\pm \sqrt{A^2+4}}{4}$$
Note that we should carefully choose the '$\pm$' from $A$ , since $|\cos x| \le 1$
We then get $x$ with the trigonometric inverse of the valid Quadratic Solutions , which will have the Principal Value :
$$\theta = \arccos\left(\frac{A\pm \sqrt{A^2+4}}{4}\right)$$
Solutions :
$$x = 2 \pi n \pm \theta$$
Maybe my calculation is correct.
[[ User Sahaj & User Acccumulation have made corrections to the calculations via comments ]]