Trig+Logs

$\mathrm{sin(log(x))+2cos(3log(x))sin(2log(x))= sin(log(x))+ sin(5log(x))-sin(log(x))}$ $=\mathrm{sin(5log(x))= 0\implies 5log(x)=kπ }$ where $(k∈Z)$ $x=e^{\frac{kπ}{5}}$ Then solution is $\color{#20A900}\boxed{x=e^{\frac{π}{5}}=1.8744}$ Here smallest possible $n=1$ [as $x>1$ and $x≠0$ ]

Let $\theta = \ln x$ . Then we have:

$\begin{aligned} \sin \theta + 2 \cos 3\theta \sin 2\theta & = 0 \\ \sin \theta + 2(4 \cos^3 \theta - 3\cos \theta) \cdot 2 \sin \theta \cos \theta & = 0 & \small \blue{\text{Since } \theta = \ln x > 0 \implies \sin \theta \ne 0} \\ 16 \cos^4 \theta - 12 \cos^2 \theta + 1 & = 0 & \small \blue{\text{Divide both sides by }\sin \theta} \\ \cos^2 \theta & = \frac {3\pm \sqrt 5}8 \\ \cos \theta & = \pm \frac {\sqrt 5 \pm 1}4 \\ \theta = \ln x & = \left(n + \frac k5\right)\pi & \small \blue{\text{where }n \in \mathbb Z, k = 1,2,3,4} \end{aligned}$

For the smallest $x>1$ , $\implies \theta > \ln 1 = 0$ or $\left(n+\dfrac k5\right)\pu > 0 \implies n = 0, k = 1$ and $\ln x = \dfrac \pi 5 \implies x = e^\frac \pi 5 \approx \boxed{1.87}$ .

pic of the solution