What is the general solution for 0=-9*sin(2x) ?

The general solution for 0=-9*sin(2x) is x=pin,x= pi/2+pin

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0=-9*sin(2x)

0 = - 9 \cdot sin (2 x)

x = πn, x = \frac{π}{2} + πn

Degrees

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

Solution steps

0 = - 9 sin (2 x)

Switch sides

- 9 sin (2 x) = 0

Divide both sides by

- 9

- 9 sin (2 x) = 0

Divide both sides by

- 9

\frac{- 9 sin ( 2 x )}{- 9} = \frac{0}{- 9}

Simplify

sin (2 x) = 0

sin (2 x) = 0

General solutions for

sin (2 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}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

sec^{2} (x) = 4 tan^{2} (x)

5 sin (x) = - 4

- π sin (π x) = 0

7 tan (4 x + 199) = 8 sin (4 x + 199)

4 sin^{2} (x) + 4 cos (x) - 5 = 0, 0 \leq x \leq 2 π

What is the general solution for 0=-9*sin(2x) ?
The general solution for 0=-9*sin(2x) is x=pin,x= pi/2+pin