Solve the trigonometric equation $4\tan(3x)=-3\tan(4x)$.
Is it possible to solve this equation without going to $\tan(x)$ or to the angle $x/2$. There the equations of very high degrees are obtained. And it seems that the roots are not good, that is, a cubic equation with bad coefficients arises. Help me, please.
Best Answer
If you expand the tangents and simplify, you should have $$4 \tan (3 x)+3 \tan (4 x)=\frac{(7 \sin (7 x)-\sin (x)) \sec (x) \sec (4 x)}{4 \cos (2 x)-2}$$ Now, with $t=\sin(x)$ $$7 \sin (7 x)-\sin (x)=-8\sin(x) \left(56 t^6-98 t^4+49 t^2-6\right)$$ Solving the cubic in $y=t^2$, we have $\Delta=111524$; so three real roots.
Using the trigonometric (!!) method for such a case
$$y_k=\frac{\sqrt{7}}{12} \left(\sqrt{7}+2 \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(-\frac{38}{49 \sqrt{7}}\right)\right)\right)\right)\quad \quad [k=0,1,2]$$ All the $y_k$ are positive.