Solving $\sin(3x +28^\circ)= \cos(2x-13^\circ)$ 
Solve the following equation :
  $$\sin(3x +28^\circ)= \cos(2x-13^\circ), x \in [0^\circ, 360^\circ]$$

Solution:
$$3x+28 +2x -13 =90$$
$$x=15$$
$$3x+28 -2x +13 = 90$$
$$x= 49$$
But the value $$x = 87$$ also satisfies the equation. I do not know how it comes.
 A: Note: Using $\cos(x)=\sin(90-x)$ (since you are working in degrees) your equation becomes:
$$\sin(3x +28)= \sin(90-2x+13), x \in [0, 360]$$
Now, if $\sin(a)=\sin(b)$ for $a,b \in [0,360]$ you must have $a=b+360 k$ or $a+b=180 \cdot k$.
A: Alternatively:
$$\sin(3x +28^\circ)= \cos(2x-13^\circ) \Rightarrow \\
\sin(3x +28^\circ)= \sin(90^\circ -(2x-13^\circ)) \Rightarrow \\
\sin(3x +28^\circ)-\sin(103^\circ-2x)=0 \Rightarrow \\
2\cos \frac{3x+28^\circ+103^\circ-2x}{2}\sin \frac{3x+28^\circ-(103^\circ -2x)}{2}=0 \Rightarrow \\
\cos \frac{x+131^\circ}{2}\cdot \sin \frac{5x-75^\circ}{2}=0 \Rightarrow \\
1) \ \cos \frac{x+131^\circ}{2}=0 \Rightarrow \frac{x+131^\circ}{2}=\frac{\pi}{2}+\pi n,n\in Z \Rightarrow \\
x=\pi+2\pi n-131^\circ, n\in Z \Rightarrow \\
0^\circ\le \pi+2\pi n-131^\circ\le 360^\circ \Rightarrow \\
0^\circ\le 180^\circ+360^\circ n-131^\circ\le 360^\circ \Rightarrow \\
0^\circ\le 360^\circ n+49^\circ\le 360^\circ \Rightarrow \\
\color{red}{x_1}=360^\circ \cdot 0+49^\circ =\color{red}{49^\circ}.\\
2) \ \sin\frac{5x-75^\circ}{2}=0 \Rightarrow \frac{5x-75^\circ}{2}=\pi k,k\in Z \Rightarrow \\
x=\frac{2\pi k}{5}+15^\circ, k\in Z \Rightarrow \\
0^\circ\le 72^\circ k+15^\circ\le 360^\circ \Rightarrow \\
\color{red}{x_2}=72^\circ \cdot 0+15^\circ =\color{red}{15^\circ};\\
\color{red}{x_3}=72^\circ \cdot 1+15^\circ =\color{red}{87^\circ};\\
\color{red}{x_4}=72^\circ \cdot 2+15^\circ =\color{red}{159^\circ};\\
\color{red}{x_5}=72^\circ \cdot 3+15^\circ =\color{red}{231^\circ};\\
\color{red}{x_6}=72^\circ \cdot 4+15^\circ =\color{red}{303^\circ}.
$$
