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 ! :)
 A: Hint:
Let $t=\sin^2 x$, so the given inequality is equivalent to $-4t^2+3t>\frac14$ or
\begin{align*}
16t^2-12t+1&<0\\
\left(4t-\frac32\right)^2-\frac54&<0
\end{align*}
Then $$-\frac{\sqrt{5}}2<4t-\frac32<\frac{\sqrt{5}}2\qquad\text{or}\qquad \frac{3-\sqrt5}8<t<\frac{3+\sqrt5}8$$
So we need to solve
$$\frac{3-\sqrt5}8<\sin^2 x<\frac{3+\sqrt5}8$$
A: 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.
A: $$sin(x)sin(3x) = sin(2x-x) sin(2x+x) < \frac{1}{4}$$
$$ \text{or, } \frac{cos(x) - cos(2x)}{2} < \frac{1}{4}$$
$$ \text{or, } {cos(x) - cos(2x)} - \frac{1}{2} < 0$$
$$ \text{or, } {cos(x) - 2cos^2(x) + 1} - \frac{1}{2} < 0$$
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. 
A: ‎$\require{cancel}$‎
\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 $\color{blue}{306^\circ<x<342^\circ}$.
