Solve s(pi/2+x)-sin(pi/2+x)=0

$$s ( \frac { \pi } { 2 } + x ) - \sin ( \frac { \pi } { 2 } + x ) = 0$$

Solve for s

$$\left\{\begin{matrix}s=\frac{2\sin(\frac{2x+\pi }{2})}{2x+\pi }\text{, }&x\neq -\frac{\pi }{2}\\s\in \mathrm{R}\text{, }&x=-\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2s\left(\frac{\pi }{2}+x\right)-2\sin(\frac{\pi }{2}+x)=0$$

Use the distributive property to multiply 2s by \frac{\pi }{2}+x.

$$2s\times \frac{\pi }{2}+2sx-2\sin(\frac{\pi }{2}+x)=0$$

Express 2\times \frac{\pi }{2} as a single fraction.

$$\frac{2\pi }{2}s+2sx-2\sin(\frac{\pi }{2}+x)=0$$

Cancel out 2 and 2.

$$\pi s+2sx-2\sin(\frac{\pi }{2}+x)=0$$

Add 2\sin(\frac{\pi }{2}+x) to both sides.

$$\pi s+2sx=0+2\sin(\frac{\pi }{2}+x)$$

Anything plus zero gives itself.

$$\pi s+2sx=2\sin(\frac{\pi }{2}+x)$$

Combine all terms containing s.

$$\left(\pi +2x\right)s=2\sin(\frac{\pi }{2}+x)$$

The equation is in standard form.

$$\left(2x+\pi \right)s=2\sin(x+\frac{\pi }{2})$$

Divide both sides by \pi +2x.

$$\frac{\left(2x+\pi \right)s}{2x+\pi }=\frac{2\sin(\frac{2x+\pi }{2})}{2x+\pi }$$

Dividing by \pi +2x undoes the multiplication by \pi +2x.

$$s=\frac{2\sin(\frac{2x+\pi }{2})}{2x+\pi }$$