What is the general solution for 2cot^2(t)sin(t)-cot^2(t)=0 ?

The general solution for 2cot^2(t)sin(t)-cot^2(t)=0 is t= pi/2+pin,t= pi/6+2pin,t=(5pi)/6+2pin

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2cot^2(t)sin(t)-cot^2(t)=0

2 cot^{2} (t) sin (t) - cot^{2} (t) = 0

t = \frac{π}{2} + πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Solution steps

2 cot^{2} (t) sin (t) - cot^{2} (t) = 0

Factor

2 cot^{2} (t) sin (t) - cot^{2} (t) : cot^{2} (t) (2 sin (t) - 1)

cot^{2} (t) (2 sin (t) - 1) = 0

Solving each part separately

cot^{2} (t) = 0 or 2 sin (t) - 1 = 0

cot^{2} (t) = 0 : t = \frac{π}{2} + πn

2 sin (t) - 1 = 0 : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

7 cos (x) = - 1

2 sin^{2} (θ) + cos (θ) - 2 = - 1

sin^{2} (x) - 6 cos (x) + 6 = 0

cos (x) = \frac{14.3}{19}

- 12 sin (x) = 0

What is the general solution for 2cot^2(t)sin(t)-cot^2(t)=0 ?
The general solution for 2cot^2(t)sin(t)-cot^2(t)=0 is t= pi/2+pin,t= pi/6+2pin,t=(5pi)/6+2pin