# The solutions of the equation $\sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}}$ are?

I tried this:

$$\sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}}$$

$$2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}}$$

$$4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}}$$

$$\sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}}$$

Here, I tried to set $$\sin x = t$$

$$t(1-t^2) = \frac{2}{3\sqrt{3}},$$ but I don't know to resolve this.

HINT

To begin with, notice that $$\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$$. Hence it results that: \begin{align*} \sin(x) + \sin(3x) = \frac{8}{3\sqrt{3}} & \Longleftrightarrow 4\sin(x) - 4\sin^{3}(x) = \frac{8}{3\sqrt{3}}\\\\ & \Longleftrightarrow \sin(x) - \sin^{3}(x) = \frac{2\sqrt{3}}{9} \end{align*}

By inspection, one concludes that \begin{align*} \sin(x) = \frac{1}{\sqrt{3}} \end{align*}

satisfies the resulting equation.

From then on, you can factor the cubic equation to obtain a quadratic which is easy to deal with.

Can you take it from here?

• Yes, I did it. Thanks a lot!! May 2, 2022 at 1:11
• @slowlyn you are welcome! I am glad to help. May 2, 2022 at 1:12

As you obtained, $$\displaystyle t-t^3 = \frac{2}{3\sqrt{3}}~, \text {where } t = \sin x$$

$$\displaystyle t - t^3 = \frac{1}{\sqrt3} - \frac{1}{3 \sqrt3} \implies \left(\frac{1}{\sqrt3}\right)^3 - t^3 = \left(\frac{1}{\sqrt3} - t\right)$$

Now using the fact that $$a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$$

One of the obvious solution is $$~t = \dfrac{1}{\sqrt3}$$.

If $$~t \ne \dfrac{1}{\sqrt3},$$ we have $$~\displaystyle t^2 + \frac{1}{3} + \frac{t}{\sqrt3} = 1$$

i.e. $$~ \displaystyle \left(t + \frac{1}{2 \sqrt3}\right)^2 = \frac{3}{4}~$$. That gives $$\displaystyle t = \frac{1}{\sqrt3}, - \frac{2}{\sqrt3}$$

As $$~- \dfrac{2}{\sqrt3} \lt - 1$$, that is not a valid solution. So the only solution is $$\sin x = \dfrac{1}{\sqrt3}$$.