Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$

Question

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 ! :)

Answer 1

‎$\require{cancel}$‎ \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}

Answer 2

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

Answer 3

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:

$$x<18^\circ$$ $$18^\circ72^\circ$$

And see which region(s) satisfy the inequality.

Answer 4

$$sin(x)sin(3x) = sin(2x-x) sin(2x+x) < \frac{1}{4}$$ $$ \text{or, } \frac{cos(x) - cos(2x)}{2} < \frac{1}{4}$$ $$ \text{or, } {cos(x) - cos(2x)} - \frac{1}{2} < 0$$ $$ \text{or, } {cos(x) - 2cos^2(x) + 1} - \frac{1}{2} < 0$$ 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.