Trigonometry Challenge (Find x)

[anemone]

Solve $\large 3^{3\cos x(1+\sin^2 x)}-3^{\cos x(4-\sin^2 x)}=6\cos 3x$.

 

[anemone]

Hint:

Spoiler

Try to rewrite the given equation such that one side is always greater than or equal to zero, and the other side is always less than or equal to zero...

 

[anemone]

Solution of other:

Spoiler

$ 3^{3\cos x(1+\sin^2 x)}-3^{\cos x(4-\sin^2 x)}=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}\left(1-\dfrac{3^{\cos x(4-\sin^2 x)}}{3^{3\cos x(1+\sin^2 x)}}\right)=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{(4\cos x-\cos x\sin^2 x)-(3\cos x+3\cos x\sin^2 x)})=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{(4\cos x-\cos x(1-\cos^2 x))-(3\cos x+3\cos x(1-\cos^2 x))})=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{(4\cos x-\cos x+\cos^3 x)-(3\cos x+3\cos x-3\cos^3 x)})=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{4\cos^3 x-3\cos x})=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{\cos 3x})=6\cos 3x$

$ 3^{3\cos x(1+\sin^2 x)}(1-3^{\cos 3x})(1-3^{\cos 3x})=6\cos 3x(1-3^{\cos 3x})$

$ \dfrac{3^{3\cos x(1+\sin^2 x)}(1-3^{\cos 3x})^2}{6}=\cos 3x(1-3^{\cos 3x})$ (*)Since $y=f(x)(1-3^{f(x)})\le 0$ for all $x$, the RHS of the equation above (*) is always less than or zero to zero for all $x$

But notice that on the LHS of the equation (*), we have

1. A squared term on the LHS that is always greater than or equal to zero: $(1-3^{\cos 3x})^2\ge 0$,

2. $\dfrac{3^{3\cos x(1+\sin^2 x)}}{6}$ that is always greater than zero.

That means the equation holds iff both sides equal to zero, and that happens when:

$\cos 3x=0$

$3x=\dfrac{\pi}{2}+n\pi$.

$\therefore x=\dfrac{\pi}{6}+\dfrac{n\pi}{3}$ for integer $n$.