Find all solutions to the equation:$3^{\sin x} \cos x-3^{\cos x} \sin x=0$ Find all solutions to the equation:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0$
My attempt:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0\implies 3^{\sin x} \cos x=3^{\cos x} \sin x\implies3^{\sin x-\cos x}=\tan x.$
I have found that $x=45^{\circ}$ or $225^{\circ}$ are satisfying the equation.
But,I don't know if these are the only solutions.
If these are the only solutions,then how to prove it?
Any guidance is welcome!
Thank you!
 A: Assume $x \in [0,2\pi]$. Rewrite it : $3^{\cos x}\sin x = 3^{\sin x}\cos x$. If $\sin x > 0 \implies \cos x > 0$ and the same for the case negative sign. Thus you might consider $x \in (0,\pi/2)\cup(\pi, 3\pi/2)$. For the first case, rewrite further: $\dfrac{3^{\sin x}}{\sin x}= \dfrac{3^{\cos x}}{\cos x}$. Consider $f(t) = \dfrac{3^t}{t}$. We have: $f'(t) = \dfrac{t3^t\ln 3 - 3^t}{t^2}= \dfrac{3^t(t\ln 3 - 1)}{t^2}= 0 \implies t = \dfrac{1}{\ln 3}$. Observe that $\dfrac{1}{\ln 3} = 0.91024 > 0.78539 =\dfrac{\pi}{4} > 0$. So if $0 < t < \pi/4 \implies f'(t) < 0$, and in this interval $0 <\sin x < \cos x\implies f(\sin x) > f(\cos x)\implies \dfrac{3^{\sin x}}{\sin x} > \dfrac{3^{\cos x}}{\cos x}$. Thus the equation has no solution. If $\pi/4 < t < \dfrac{1}{\ln 3}\implies f'(t) < 0$ and on this interval $0 < \cos x < \sin x\implies f(\cos x) > f(\sin x) \implies \dfrac{3^{\cos x}}{\cos x} > \dfrac{3^{\sin x}}{\sin x}$. Thus the equation also has no solution on this interval. At $x=\pi/4$, the equation holds and therefore $x = \pi/4$ is a solution. If $\dfrac{1}{\ln 3} < t \le \pi/2\implies f'(t) > 0$ and on this interval $\sin x > \cos x \implies f(\sin x) > f(\cos x)\implies \dfrac{3^{\sin x}}{\sin x} > \dfrac{3^{\cos x}}{\cos x}$. So the equation has no solution on this interval. At $t = \dfrac{1}{\ln 3} = 0.91024$, $\dfrac{3^{\sin(0.91024)}}{\sin(0.91024)}= 3.01525 < 3.19806 = \dfrac{3^{\cos(0.91024)}}{\cos(0.91024)}$. Thus $x = \dfrac{1}{\ln 3}$ is not a solution.Thus there is only one solution on the interval $(0,\pi/2)$ and that solution is $x = \pi/4 = 45^{\circ}$. Similarly you can show that the other solution is $x = 5\pi/4 = 225^{\circ}$, and this completes the proof that there are only two named solutions.
A: You are pretty much at the solution.
You have 3^{Sin x - Cos x} = Sin x/Cos x.
As you noticed at x = pi/4 or 5pi/4, both sides become equal. In fact this would hold for all x where sin x = cos x as 3^(0) = 1 = RHS.
