Solve n(2/3pi+x)-sqrt{3}cos^2x=sinxcos(x-pi/6)

$$n ( \frac { 2 } { 3 } \pi + x ) - \sqrt { 3 } \cos ^ { 2 } x = \sin x \cos ( x - \frac { \pi } { 6 } )$$

Solve for n

$$n=\frac{3\left(4\sqrt{3}\left(\cos(x)\right)^{2}+2\left(\sin(x)\right)^{2}+\sqrt{3}\sin(2x)\right)}{4\left(3x+2\pi \right)}$$

$$x\neq -\frac{2\pi }{3}$$

Steps for Solving Linear Equation

Use the distributive property to multiply n by \frac{2}{3}\pi +x.

$$\frac{2}{3}n\pi +nx-\sqrt{3}\left(\cos(x)\right)^{2}=\sin(x)\cos(x-\frac{\pi }{6})$$

Add \sqrt{3}\left(\cos(x)\right)^{2} to both sides.

$$\frac{2}{3}n\pi +nx=\sin(x)\cos(x-\frac{\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}$$

Combine all terms containing n.

$$\left(\frac{2}{3}\pi +x\right)n=\sin(x)\cos(x-\frac{\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}$$

The equation is in standard form.

$$\left(x+\frac{2\pi }{3}\right)n=\sin(x)\cos(x-\frac{\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}$$

Divide both sides by \frac{2}{3}\pi +x.

$$\frac{\left(x+\frac{2\pi }{3}\right)n}{x+\frac{2\pi }{3}}=\frac{\sin(x)\cos(\frac{6x-\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}}{x+\frac{2\pi }{3}}$$

Dividing by \frac{2}{3}\pi +x undoes the multiplication by \frac{2}{3}\pi +x.

$$n=\frac{\sin(x)\cos(\frac{6x-\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}}{x+\frac{2\pi }{3}}$$

Divide \cos(\frac{6x-\pi }{6})\sin(x)+\left(\cos(x)\right)^{2}\sqrt{3} by \frac{2}{3}\pi +x.

$$n=\frac{3\left(\sin(x)\cos(\frac{6x-\pi }{6})+\sqrt{3}\left(\cos(x)\right)^{2}\right)}{3x+2\pi }$$