The number of solutions for $\frac{\sin 2x}{2}+1=\sin x+\cos x$ in the interval $(0,2\pi ]$ is .

Let's solve the trigonometric equation step by step using the factorization method.

Given: $\frac{\sin 2x}{2}+1=\sin x+\cos x$
$\Rightarrow \frac{2\sin x\cos x}{2}+1=\sin x+\cos x$
$\Rightarrow \sin x\cos x+1=\sin x+\cos x$

Step 1: Transform the equation into product of linear factors by bringing the RHS term to LHS and make the RHS as zero.
$\Rightarrow \sin x\cos x+1-\sin x-\cos x=0$
$\Rightarrow \sin x\cos x-\sin x+1-\cos x=0$

(taking $\sin x$ common from first two terms)
$\Rightarrow \sin x(\cos x-1)+1-\cos x=0$
(taking $-1$ common from last two terms)
$\Rightarrow \sin x(\cos x-1)-1(-1+\cos x)=0$
$\Rightarrow \sin x(\cos x-1)-1(\cos x-1)=0$

(taking $(\cos x-1)$ common)
$\Rightarrow (\cos x-1)(\sin x-1)=0$

Step 2: Equate each factor to zero.
$\therefore$ Either $\cos x-1=0$ or $\sin x-1=0$

Step 3: Solve each equation individually.
For $\cos x-1=0,$ i.e., $\cos x=1,x=2\pi$ in the interval $(0,2\pi ].$

For $\sin x-1=0,$ i.e., $\sin x=1,x=\frac{\pi }{2}$ in the interval $(0,2\pi ].$

Step 4: Take union of all these solutions.
$\therefore$ The complete solution is $x=\{\frac{\pi }{2},2\pi \}$ in the interval $(0,2\pi ].$

Hence, the number of solutions is two in the given interval.