What is the general solution for -7sin(7x)=0 ?

The general solution for -7sin(7x)=0 is x=(2pin)/7 ,x= pi/7+(2pin)/7

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-7sin(7x)=0

- 7 sin (7 x) = 0

x = \frac{2 πn}{7}, x = \frac{π}{7} + \frac{2 πn}{7}

Degrees

x = 0^{\circ} + 51.42857 \dots^{\circ} n, x = 25.71428 \dots^{\circ} + 51.42857 \dots^{\circ} n

Solution steps

- 7 sin (7 x) = 0

Divide both sides by

- 7

- 7 sin (7 x) = 0

Divide both sides by

- 7

\frac{- 7 sin ( 7 x )}{- 7} = \frac{0}{- 7}

Simplify

sin (7 x) = 0

sin (7 x) = 0

General solutions for

sin (7 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}

7 x = 0 + 2 πn, 7 x = π + 2 πn

7 x = 0 + 2 πn, 7 x = π + 2 πn

Solve

7 x = 0 + 2 πn : x = \frac{2 πn}{7}

7 x = 0 + 2 πn

0 + 2 πn = 2 πn

7 x = 2 πn

Divide both sides by

7

7 x = 2 πn

Divide both sides by

7

\frac{7 x}{7} = \frac{2 πn}{7}

Simplify

x = \frac{2 πn}{7}

x = \frac{2 πn}{7}

Solve

7 x = π + 2 πn : x = \frac{π}{7} + \frac{2 πn}{7}

7 x = π + 2 πn

Divide both sides by

7

7 x = π + 2 πn

Divide both sides by

7

\frac{7 x}{7} = \frac{π}{7} + \frac{2 πn}{7}

Simplify

x = \frac{π}{7} + \frac{2 πn}{7}

x = \frac{π}{7} + \frac{2 πn}{7}

x = \frac{2 πn}{7}, x = \frac{π}{7} + \frac{2 πn}{7}

5 cos (x) + 3 = - cos (x)

9 tan (x) sin (x) = - 5 sin (x)

cos (2 x) = 0, 0 \leq x \leq 2 π

cot^{2} (θ) - csc (θ) = 1

cos (\frac{x}{2}) = - \frac{3}{2}

What is the general solution for -7sin(7x)=0 ?
The general solution for -7sin(7x)=0 is x=(2pin)/7 ,x= pi/7+(2pin)/7