Solve cosx+sinx=2w

$$\cos x + \sin x = 2 w$$

Solve for w (complex solution)

$$w=\left(\frac{1}{4}-\frac{1}{4}i\right)e^{ix}+\left(\frac{1}{4}+\frac{1}{4}i\right)e^{-ix}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2w=\cos(x)+\sin(x)$$

The equation is in standard form.

$$2w=\sin(x)+\cos(x)$$

Divide both sides by 2.

$$\frac{2w}{2}=\frac{\sin(x)+\cos(x)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$w=\frac{\sin(x)+\cos(x)}{2}$$

Solve for w

$$w=\frac{\sin(x)+\cos(x)}{2}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2w=\cos(x)+\sin(x)$$

The equation is in standard form.

$$2w=\sin(x)+\cos(x)$$

Divide both sides by 2.

$$\frac{2w}{2}=\frac{\sin(x)+\cos(x)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$w=\frac{\sin(x)+\cos(x)}{2}$$

Solve for x (complex solution)

$$x=-i\ln(\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt{4w^{2}-2}+\left(1+i\right)w)+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$x=-i\ln(\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt{4w^{2}-2}+\left(1+i\right)w)+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$