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 ! :)

You are on the correct path.
Now you need to solve the following equation:
$$\sin^2(x)\cdot \big(3-4\sin^2(x)\big) =\frac{1}{4}$$
$$4\sin^2(x)\cdot\big(3-4\sin^2(x)\big)-1=0$$
$$12\sin^2x-16\sin^4x-1=0$$
$$16\sin^4x-12\sin^2x+1=0$$
So we get by solving, $$\sin^2x=\frac{12\pm \sqrt{144-4\cdot 16}}{2\cdot 16}=\frac{12\pm \sqrt{80}}{32}=\frac{6\pm 2\sqrt{5}}{16}=\left(\frac{\sqrt5 \pm 1}{4}\right)^2=(\sin 72^\circ)^2 \text{or} (\sin 18^\circ)^2$$
So the $4$ roots are $x=72^\circ$,$x=72^\circ$,$x=18^\circ$ and $x=18^\circ$.
Now observe that, to check the inequality, you have to check for three regions:
$$x<18^\circ$$ $$18^\circ<x<72^\circ$$ $$x>72^\circ$$ 
And see which region(s) satisfy the inequality.
