Solve f^prime(x)=2x*a*cos2x-1/2cosx

$$f ^ { \prime } ( x ) = 2 x \cdot a \cdot \cos 2 x - \frac { 1 } { 2 } \cos x$$

Solve for a

$$a=\frac{\cos(x)}{4x\cos(2x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

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

$$2xa\cos(2x)-\frac{1}{2}\cos(x)=\frac{\mathrm{d}}{\mathrm{d}x}(f)x$$

Add \frac{1}{2}\cos(x) to both sides.

$$2xa\cos(2x)=\frac{\mathrm{d}}{\mathrm{d}x}(f)x+\frac{1}{2}\cos(x)$$

The equation is in standard form.

$$2x\cos(2x)a=\frac{\cos(x)}{2}$$

Divide both sides by 2x\cos(2x).

$$\frac{2x\cos(2x)a}{2x\cos(2x)}=\frac{\cos(x)}{2\times 2x\cos(2x)}$$

Dividing by 2x\cos(2x) undoes the multiplication by 2x\cos(2x).

$$a=\frac{\cos(x)}{2\times 2x\cos(2x)}$$

Divide \frac{\cos(x)}{2} by 2x\cos(2x).

$$a=\frac{\cos(x)}{4x\cos(2x)}$$

Solve for f (complex solution)

$$f\in \mathrm{C}$$

$$a=\frac{\cos(x)}{4x\cos(2x)}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{4}\text{ and }x\neq 0$$

Solve for f

$$f\in \mathrm{R}$$

$$a=\frac{\cos(x)}{4x\cos(2x)}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{4}\text{ and }x\neq 0$$