Solve (-sin^2x)=operatorname{ctg}x-T

$$( - \sin ^ { 2 } x ) = \operatorname { ctg } x - T$$

Solve for T

$$T=\frac{\left(\sin(x)\right)^{3}+\cos(x)}{\sin(x)}$$

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

Steps for Solving Linear Equation

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

$$\cot(x)-T=-\left(\sin(x)\right)^{2}$$

Subtract \cot(x) from both sides.

$$-T=-\left(\sin(x)\right)^{2}-\cot(x)$$

Divide both sides by -1.

$$\frac{-T}{-1}=-\frac{\frac{\left(\sin(x)\right)^{3}+\cos(x)}{\sin(x)}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$T=-\frac{\frac{\left(\sin(x)\right)^{3}+\cos(x)}{\sin(x)}}{-1}$$

Divide -\frac{\cos(x)+\left(\sin(x)\right)^{3}}{\sin(x)} by -1.

$$T=\frac{\left(\sin(x)\right)^{3}+\cos(x)}{\sin(x)}$$