Solve 1-2sin^2x+asinx=a

$$1 - 2 \sin ^ { 2 } x + a \sin x = a$$

Solve for a

$$a=-\frac{\cos(2x)}{\sin(x)-1}$$

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

Steps for Solving Linear Equation

Subtract a from both sides.

$$1-2\left(\sin(x)\right)^{2}+a\sin(x)-a=0$$

Subtract 1-2\left(\sin(x)\right)^{2} from both sides. Anything subtracted from zero gives its negation.

$$a\sin(x)-a=-\left(1-2\left(\sin(x)\right)^{2}\right)$$

To find the opposite of 1-2\left(\sin(x)\right)^{2}, find the opposite of each term.

$$a\sin(x)-a=-1+2\left(\sin(x)\right)^{2}$$

Combine all terms containing a.

$$\left(\sin(x)-1\right)a=-1+2\left(\sin(x)\right)^{2}$$

The equation is in standard form.

$$\left(\sin(x)-1\right)a=-\cos(2x)$$

Divide both sides by \sin(x)-1.

$$\frac{\left(\sin(x)-1\right)a}{\sin(x)-1}=-\frac{\cos(2x)}{\sin(x)-1}$$

Dividing by \sin(x)-1 undoes the multiplication by \sin(x)-1.

$$a=-\frac{\cos(2x)}{\sin(x)-1}$$