cos^2t*sint-sint=0

cos^2t*sint-sint = 0     t = ?

$$\boxed{ ~\cos^2{(t)}\cdot \sin{(t)}-\sin{(t)}=0~ }\\\\ \small{\text{ $ \begin{array}{rcl} \cos^2{(t)}\cdot \sin{(t)}-\sin{(t)} &=& 0 \qquad | \qquad \boxed{~ \cos^2{(t)} = 1-\sin^2{(t)} ~}\\ \left[ 1-\sin^2{(t)} \right] \cdot \sin{(t)}-\sin{(t)} &=& 0\\ \sin{(t)}-\sin^3{(t)} - \sin{(t)} &=& 0\\ -\sin^3{(t)} &=& 0\\ \sin^3{(t)} &=& 0 \qquad | \qquad \sqrt[3]{} \\ \sin{(t)} &=& \sqrt[3]{0}\\ \sin{(t)} &=& 0 \end{array} $}}\\\\ \small{\text{ $ \begin{array}{rcl} \sin{(t)} &=& 0 \\ t &=& \arcsin{(0)} \\ t &=& 0 \pm \rm{k}\cdot 360\ensurement{^{\circ}} \end{array} $}}\\\\\\ \small{\text{ $ \begin{array}{rcl} \sin{(t)} = \sin{(180\ensurement{^{\circ}}- t)} &=& 0 \\ 180\ensurement{^{\circ}}- t &=& \arcsin{(0)} \\ 180\ensurement{^{\circ}}- t &=& 0 \\ t &=& 180\ensurement{^{\circ}} \pm \rm{k}\cdot 360\ensurement{^{\circ}} \end{array} $}}$$

$$\small{\text{ So~ $ t= 0 \pm k\cdot 180 \ensurement{^{\circ}} \qquad k = 0,1,2\cdots \in N $}}$$

cos²(t)·sin(t) - sin(t)  =  0

Factor out sin(t):

sin(t)[cos²(t) - 1]  =  0

Factor again:

sin(t)[cos(t) + 1][cos(t) - 1]  =  0

So either:

sin(t)  =  0     or     cos(t) + 1  =  0     or     cos(t) - 1  =  0

sin(t)  =  0     or     cos(t)  =  -1     or    cos(t)  =  1

Therefore:

sin(t)  =  0     --->   t  =  0° (plus multiples of 360°)   or   t  =  180° (plus multiples of 360°)

cos(t)  =  -1   --->   t  =  0° (plus multiples of 360°)

cos(t)  =  1    --->   t  =  180° (plus multiples of 360°)

So:  t  =  0° (plus multiples of 360°)   or   t  =  180° (plus multiples of 360°)