What is the general solution for -5sin(2t)=0 ?

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

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-5sin(2t)=0

- 5 sin (2 t) = 0

t = πn, t = \frac{π}{2} + πn

Degrees

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

Solution steps

- 5 sin (2 t) = 0

Divide both sides by

- 5

- 5 sin (2 t) = 0

Divide both sides by

- 5

\frac{- 5 sin ( 2 t )}{- 5} = \frac{0}{- 5}

Simplify

sin (2 t) = 0

sin (2 t) = 0

General solutions for

sin (2 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}

2 t = 0 + 2 πn, 2 t = π + 2 πn

2 t = 0 + 2 πn, 2 t = π + 2 πn

Solve

2 t = 0 + 2 πn : t = πn

2 t = 0 + 2 πn

0 + 2 πn = 2 πn

2 t = 2 πn

Divide both sides by

2

2 t = 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{2 πn}{2}

Simplify

t = πn

t = πn

Solve

2 t = π + 2 πn : t = \frac{π}{2} + πn

2 t = π + 2 πn

Divide both sides by

2

2 t = π + 2 πn

Divide both sides by

2

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

Simplify

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

t = πn, t = \frac{π}{2} + πn

(cos (x) - 1) = 0

1 + cot^{2} (a) = tan^{2} (a)

tan (α) = \frac{12}{6 3}

1 = cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

4 sin (θ) = 1

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