Solve cos^2x-operatorname{gsin}x+5=0

$$\cos ^ { 2 } x - \operatorname { gsin } x + 5 = 0$$

Solve for g

$$g=\frac{6}{\sin(x)}-\sin(x)$$

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

Steps for Solving Linear Equation

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$\left(\cos(x)\right)^{2}-g\sin(x)=-5$$

Subtract \left(\cos(x)\right)^{2} from both sides.

$$-g\sin(x)=-5-\left(\cos(x)\right)^{2}$$

The equation is in standard form.

$$\left(-\sin(x)\right)g=-\left(\cos(x)\right)^{2}-5$$

Divide both sides by -\sin(x).

$$\frac{\left(-\sin(x)\right)g}{-\sin(x)}=\frac{\left(\sin(x)\right)^{2}-6}{-\sin(x)}$$

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

$$g=\frac{\left(\sin(x)\right)^{2}-6}{-\sin(x)}$$

Divide -6+\left(\sin(x)\right)^{2} by -\sin(x).

$$g=\frac{\left(\cos(x)\right)^{2}+5}{\sin(x)}$$