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 ! :)

$$sin(x)sin(3x) = sin(2x-x) sin(2x+x) < \frac{1}{4}$$
$$ \text{or, } \frac{cos(x) - cos(2x)}{2} < \frac{1}{4}$$
$$ \text{or, } {cos(x) - cos(2x)} - \frac{1}{2} < 0$$
$$ \text{or, } {cos(x) - 2cos^2(x) + 1} - \frac{1}{2} < 0$$
Let $y = cos(x)$
$$ \text{or, } - 2y^2 + y + \frac{1}{2} < 0$$
Solving, 
$$ \frac{1-\sqrt{5}}{4} < cos(x) < \frac{1+\sqrt{5}}{4}$$
Hence,
$$\frac{\pi}{5} < x < \frac{3 \pi}{5}$$ and other $2 \pi n$ difference of it. 
