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

The general solution for 2sin^2(t)cos(t)=0 is t=2pin,t=pi+2pin,t= pi/2+2pin,t=(3pi)/2+2pin

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2sin^2(t)cos(t)=0

2 sin^{2} (t) cos (t) = 0

t = 2 πn, t = π + 2 πn, t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (t) cos (t) = 0

Solving each part separately

sin^{2} (t) = 0 or cos (t) = 0

sin^{2} (t) = 0 : t = 2 πn, t = π + 2 πn

sin^{2} (t) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (t) = 0

General solutions for

sin (t) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

General solutions for

cos (t) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

Combine all the solutions

t = 2 πn, t = π + 2 πn, t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

2 cos (θ) + sin (θ) = 0

2 sin^{2} (x) + sin (x) - 1 = 0, 0 < x < 2 π

5 sin (x) - 12 cos (x) = 13

sec (x) = 1.562641

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

What is the general solution for 2sin^2(t)cos(t)=0 ?
The general solution for 2sin^2(t)cos(t)=0 is t=2pin,t=pi+2pin,t= pi/2+2pin,t=(3pi)/2+2pin