What is the general solution for -25sin(5x)=0 ?

The general solution for -25sin(5x)=0 is x=(2pin)/5 ,x= pi/5+(2pin)/5

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-25sin(5x)=0

- 25 sin (5 x) = 0

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

Degrees

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

Solution steps

- 25 sin (5 x) = 0

Divide both sides by

- 25

- 25 sin (5 x) = 0

Divide both sides by

- 25

\frac{- 25 sin ( 5 x )}{- 25} = \frac{0}{- 25}

Simplify

sin (5 x) = 0

sin (5 x) = 0

General solutions for

sin (5 x) = 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 x = 0 + 2 πn, 5 x = π + 2 πn

5 x = 0 + 2 πn, 5 x = π + 2 πn

Solve

5 x = 0 + 2 πn : x = \frac{2 πn}{5}

5 x = 0 + 2 πn

0 + 2 πn = 2 πn

5 x = 2 πn

Divide both sides by

5

5 x = 2 πn

Divide both sides by

5

\frac{5 x}{5} = \frac{2 πn}{5}

Simplify

x = \frac{2 πn}{5}

x = \frac{2 πn}{5}

Solve

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

5 x = π + 2 πn

Divide both sides by

5

5 x = π + 2 πn

Divide both sides by

5

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

Simplify

x = \frac{π}{5} + \frac{2 πn}{5}

x = \frac{π}{5} + \frac{2 πn}{5}

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

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

sin (x) = \frac{9}{11}

sec^{2} (x) = - 4

0 = 2 sin (x) + 1

cos (θ) = \frac{- 1}{2}

What is the general solution for -25sin(5x)=0 ?
The general solution for -25sin(5x)=0 is x=(2pin)/5 ,x= pi/5+(2pin)/5