Notice $x=30^{\circ}$ is a solution since $\cos 40 ^ \circ = \sin 50 ^ \circ$ .

Now using $\sin(A+B)=\sin A\cos B+\sin B\cos A$ we get

$2\sin x \cos 40^{\circ} = \sin(x+20^{\circ})$

$2\sin x \cos 40^{\circ} = \sin x \cos 20^{\circ}+ \cos x \sin 20^{\circ}$

If $\cos x = 0$ , then $x = 90 ^ \circ + 180 ^\circ k$ , and we can check that these are not solutions. Thus, we can divide both sides by $\cos x \neq 0$ to get

$2\tan x \cos 40^{\circ} = \tan x \cos 20^{\circ}+ \sin 20^{\circ}$

$\tan x = \frac{\sin 20^{\circ}}{2\cos 40^{\circ} -\cos 20^{\circ}}$

Recall that $x = 30 ^ \circ$ was a solution to the original equation. The series of steps that we took have not changed the solution set. Since function $\tan x$ is one one on $(0,90^{\circ})$ , this implies that there is a unique solution to the final equation. Thus, the only solution is $x = 30 ^ \circ$ .

Hence answer is $\boxed{30^{\circ}}$

Here I will use the sum to product formulas to simplify the equation:

$2\sin x \cos 40^\circ=\sin x \cos 20^\circ+\cos x\sin 20^\circ \\ \sin x(2\cos 40^\circ-\cos 20^\circ)=\cos x \sin 20^\circ \\ \sin x(\cos 40^\circ+\cos 40^\circ-\cos 20^\circ)=\cos x \sin 20^\circ \\ \sin x(\cos 40^\circ-\sin 10^\circ)=\cos x \sin 20^\circ \\ \sin x(\sin 50^\circ-\sin 10^\circ)=\cos x \sin 20^\circ \\ \sin x(2\cos 30^\circ\sin 20^\circ)=\cos x \sin 20^\circ \\ 2\sin x\cos 30^\circ=\cos x$

Since $\cos x=0$ is not a solution:

$\dfrac{\sin x}{\cos x}=\dfrac{1}{2\cos 30^\circ} \\ \tan x=\dfrac{\sin 30^\circ}{\cos 30^\circ} \\ \tan x=\tan 30^\circ \\ x=30^\circ+180^\circ k$

We are looking for an acute angle, so the only solution is $x=30^\circ$ .

Disclaimer: This solution is incomplete. It only shows that $x=30^\circ$ is a solution but it did not show that it is a unique solution.

Apply product to sum formula: $2 \sin(A) \cos(B) = \sin\left(\frac{A+B}2 \right) + \sin \left( \frac{A-B}2 \right)$ .

So it becomes

$\begin{aligned} \sin(x+40^\circ) + \sin(x-40^\circ) &=& \sin(x+ 20^\circ) \\ \sin(x+40^\circ) &=& \sin(x+20^\circ) + \sin(40^\circ - x)\\ \cos(50^\circ - x) &=& \cos(70^\circ - x) + \cos(50^\circ + x) \\ \cos(50^\circ - x) &=& -\cos(110^\circ + x) - \cos(130^\circ - x) \\ \cos(50^\circ - x)+ \cos(130^\circ - x) + \cos(110^\circ + x) &=& 0 \\ \cos(50^\circ - x)+ \cos(60^\circ) + \cos(130^\circ - x) + \cos(110^\circ + x) &=& \frac12 \qquad (\ast) \\ \end{aligned}$

Now we just need to show that $\cos\left( \frac \pi9\right) + \cos\left( \frac {3\pi}9\right) + \cos\left( \frac {5 \pi}9\right) + \cos\left( \frac{7 \pi}9\right) = \frac12$ .

Let $w$ be a complex 9th root of unity. Then $w^9 = 1$ . factorizing: $w^8 + w^7 + w^6 + \ldots + w + 1 = 0$ and dividing by $w^4$ :

$w^4 + w^{-4} + w^3 + w^{-3} + w^2 + w^{-2} + w + w^{-1} = -1$ . But also $w^k + w^{-k} = 2\cos \left( \frac{2\pi k}{9}\right)$ . Then

$2\cos\left(\frac{8\pi}9 \right) + 2\cos\left(\frac{6\pi}9 \right) + 2\cos\left(\frac{4\pi}9 \right) + 2\cos\left(\frac{2\pi}9 \right) = - 1$

And finally

$\cos\left(\frac{\pi}9 \right) + \cos\left(\frac{3\pi}9 \right) + \cos\left(\frac{5\pi}9 \right) + \cos\left(\frac{7\pi}9 \right) = \frac12$

Equivalently,

$\cos(20^\circ) + \cos(60^\circ) + \cos(100^\circ) + \cos(140^\circ) = \frac12 \qquad (\ast\ast)$

This suggests that $x = \boxed{30^\circ}$ .