Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$ 
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)

One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I tried expanding $\sin(3x)$ by the regular indentity as :
$$\sin(x) \times \big(3\sin(x)-4\sin^3(x)\big) > \frac {1}{4}$$
$$\implies \sin^2(x)\times\big(3-4\sin^2(x)\big) >\frac{1}{4}$$
I do not find any way of proceeding. Wolfram Alpha gives 4 sets of answers. Do I have to observe this problem "case-by-case"? Can this question be solved without calculus ? Can anyone provide a hint to what should be done ?
Thanks in Advance ! :)

‎$\require{cancel}$‎
\begin{eqnarray}
\sin x\sin3x &>&\dfrac14\\
2\sin x\sin3x&>&\dfrac12\\
\cos2x-\cos4x&>&\dfrac12\\
\cos2x-2\cos^22x+1&>&\dfrac12~~~~~~~~~~~~~~~~~~,~~~~~\cos2x=t\\
4t^2-2t-1&<&0
\end{eqnarray}
with $\Delta=20$ so $t=\dfrac{2\pm2\sqrt{5}}{8}=\dfrac{1\pm\sqrt{5}}{4}$ are roots and then
$$\dfrac{1-\sqrt{5}}{4}<t<\dfrac{1+\sqrt{5}}{4}$$
$$-0.309=\dfrac{1-\sqrt{5}}{4}<\cos2x<\dfrac{1+\sqrt{5}}{4}=0.809$$
thus $\color{blue}{18^\circ<x<54^\circ}$ or $\color{blue}{126^\circ<x<162^\circ}$ or $\color{blue}{198^\circ<x<234^\circ}$ or $\color{blue}{306^\circ<x<342^\circ}$.
