What is the general solution for 0= 1/5 sin(5t) ?

The general solution for 0= 1/5 sin(5t) is t=(2pin)/5 ,t= pi/5+(2pin)/5

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0= 1/5 sin(5t)

0 = \frac{1}{5} sin (5 t)

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

Degrees

t = 0^{\circ} + 7 2^{\circ} n, t = 3 6^{\circ} + 7 2^{\circ} n

Solution steps

0 = \frac{1}{5} sin (5 t)

Switch sides

\frac{1}{5} sin (5 t) = 0

Multiply both sides by

5

\frac{1}{5} sin (5 t) = 0

Multiply both sides by

5

5 \cdot \frac{1}{5} sin (5 t) = 0 \cdot 5

Simplify

sin (5 t) = 0

sin (5 t) = 0

General solutions for

sin (5 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}

5 t = 0 + 2 πn, 5 t = π + 2 πn

5 t = 0 + 2 πn, 5 t = π + 2 πn

Solve

5 t = 0 + 2 πn : t = \frac{2 πn}{5}

5 t = 0 + 2 πn

0 + 2 πn = 2 πn

5 t = 2 πn

Divide both sides by

5

5 t = 2 πn

Divide both sides by

5

\frac{5 t}{5} = \frac{2 πn}{5}

Simplify

t = \frac{2 πn}{5}

t = \frac{2 πn}{5}

Solve

5 t = π + 2 πn : t = \frac{π}{5} + \frac{2 πn}{5}

5 t = π + 2 πn

Divide both sides by

5

5 t = π + 2 πn

Divide both sides by

5

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

Simplify

t = \frac{π}{5} + \frac{2 πn}{5}

t = \frac{π}{5} + \frac{2 πn}{5}

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

tan^{2} (x) - 1 = 0, 0 \leq x < 2 π

6 sin^{2} (x) - 5 sin (x) + 1 = 0

cos (θ) + 1 = sin (θ)

2 sec^{2} (x) - 2 = 0

2 cos (2 x) = - 2

What is the general solution for 0= 1/5 sin(5t) ?
The general solution for 0= 1/5 sin(5t) is t=(2pin)/5 ,t= pi/5+(2pin)/5