Solve a*cos^2(x+y)=b

$$a \cdot \cos ^ { 2 } ( x + y ) = b$$

Solve for a

$$\left\{\begin{matrix}a=\frac{b}{\left(\cos(x+y)\right)^{2}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}-y+\frac{\pi }{2}\\a\in \mathrm{R}\text{, }&b=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}-y+\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$\left(\cos(x+y)\right)^{2}a=b$$

Divide both sides by \left(\cos(x+y)\right)^{2}.

$$\frac{\left(\cos(x+y)\right)^{2}a}{\left(\cos(x+y)\right)^{2}}=\frac{b}{\left(\cos(x+y)\right)^{2}}$$

Dividing by \left(\cos(x+y)\right)^{2} undoes the multiplication by \left(\cos(x+y)\right)^{2}.

$$a=\frac{b}{\left(\cos(x+y)\right)^{2}}$$

Solve for b

$$b=a\left(\cos(x+y)\right)^{2}$$