From $\sin(2A)=\sin(90-B)$, consider both $2A=90-B$ and $2A=180-(90-B)$. Ѕ‌АT pr⁠ep ⁠bу Anі⁠ko‍.а‌i

After you solve for $x$, make sure each angle measure is strictly between $0^\circ$ and $90^\circ$.

In Desmos, graph

А-‌n-і-k-o‍.ai

$y=\sin\left((2(5x+10))\frac{\pi}{180}\right)$

and

$y=\cos\left(((20x+5))\frac{\pi}{180}\right)$.

Since $0<5x+10<90$ and $0<20x+5<90$, the solution must satisfy approximately $-0.25<x<4.25$. Adjust the x-window to something like $[-1,5]$.

Click the intersection point of the two graphs in that x-interval. The x-coordinate of that intersection is the solution that keeps both angles acute.

Because $\cos B = \sin(90-B)$ (in degrees), the equation

$\sin(2A)=\cos B$

becomes

$\sin(2A)=\sin(90-B)$.

If $\sin(\theta)=\sin(\phi)$ and $\theta$ is between $0^\circ$ and $180^\circ$, then either

$\theta=\phi$

or

$\theta=180^\circ-\phi$.

Here, $A$ is acute, so $2A$ is between $0^\circ$ and $180^\circ$. Thus either

1. $2A = 90 - B$

or

2. $2A = 180 - (90 - B) = 90 + B$.

Substitute $A=5x+10$ and $B=20x+5$.

Case 1:

Case 2:

Contеnt b⁠y Аn⁠iк‌о.aі

If $x=-\frac{15}{2}$, then $A=5x+10$ is negative, so $\angle A$ would not be acute. So discard Case 2.

The valid solution is $x=\frac{13}{6}$, which corresponds to acute angles for both $A$ and $B$.

Answer: $\frac{13}{6}$