Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is $3^{\sin 2x+\cos 2x}=(\sqrt{3}{)}^{\sin 2x}$. This can be rewritten as $3^{\sin 2x+\cos 2x}=3^{\frac{1}{2}\sin 2x}$. Equating the exponents, we get $\sin 2x+\cos 2x=\frac{1}{2}\sin 2x$, which simplifies to $\frac{1}{2}\sin 2x+\cos 2x=0$. Dividing by $\cos 2x$ (assuming $\cos 2x\ne 0$, otherwise $\sin 2x=0$ which doesn't satisfy the equation), we get $\frac{1}{2}\tan 2x=-1$, so $\tan 2x=-2$. Let $\alpha =\arctan (-2)$. Since $\tan \alpha =-2$, we know $\alpha \in (-\frac{\pi }{2},0)$. The general solution is $2x=n\pi +\alpha$, where $n\in \mathbb{Z}$. Thus, $x=\frac{n\pi }{2}+\frac{\alpha }{2}$. We need to find the number of solutions in $[0,3\pi ]$. So, $0\le \frac{n\pi }{2}+\frac{\alpha }{2}\le 3\pi$. This gives $-\frac{\alpha }{2}\le \frac{n\pi }{2}\le 3\pi -\frac{\alpha }{2}$. Since $\alpha \in (-\frac{\pi }{2},0)$, we have $0<-\frac{\alpha }{2}<\frac{\pi }{4}$. Therefore, $0<\frac{n\pi }{2}<3\pi +\frac{\pi }{4}$, which simplifies to $0<n<6+\frac{1}{2}=6.5$. The possible integer values for $n$ are $1,2,3,4,5,6$. Each of these values gives a distinct solution in the given interval. Hence, there are 6 solutions.