Solve ksin(x+pi/3)=k-1/2

$$k \sin ( x + \frac { \pi } { 3 } ) = k - \frac { 1 } { 2 }$$

Solve for k

$$k=-\frac{1}{2\left(\sin(\frac{3x+\pi }{3})-1\right)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=2\pi n_{1}+\frac{\pi }{6}$$

Steps for Solving Linear Equation

Subtract k from both sides.

$$k\sin(x+\frac{\pi }{3})-k=-\frac{1}{2}$$

Combine all terms containing k.

$$\left(\sin(x+\frac{\pi }{3})-1\right)k=-\frac{1}{2}$$

Divide both sides by \sin(x+\frac{1}{3}\pi )-1.

$$\frac{\left(\sin(x+\frac{\pi }{3})-1\right)k}{\sin(x+\frac{\pi }{3})-1}=-\frac{\frac{1}{2}}{\sin(x+\frac{\pi }{3})-1}$$

Dividing by \sin(x+\frac{1}{3}\pi )-1 undoes the multiplication by \sin(x+\frac{1}{3}\pi )-1.

$$k=-\frac{\frac{1}{2}}{\sin(x+\frac{\pi }{3})-1}$$

Divide -\frac{1}{2} by \sin(x+\frac{1}{3}\pi )-1.

$$k=-\frac{1}{2\left(\sin(\frac{3x+\pi }{3})-1\right)}$$

Solve for x

$$x=\frac{6\pi n_{1}-3\arcsin(\frac{2-\frac{1}{k}}{2})+2\pi }{3}\text{, }n_{1}\in \mathrm{Z}$$

$$x=\frac{6\pi n_{2}+3\arcsin(\frac{2-\frac{1}{k}}{2})-\pi }{3}\text{, }n_{2}\in \mathrm{Z}\text{, }k\geq \frac{1}{4}$$