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

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

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-sin(2x)(2)=0

- sin (2 x) (2) = 0

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

- sin (2 x) (2) = 0

Divide both sides by

- 2

- sin (2 x) \cdot 2 = 0

Divide both sides by

- 2

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

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

cos (5 x - 21) = sin (13 - 3 x)

sin (θ) = \frac{5}{3}

cos (2 θ) + 14 sin^{2} (θ) = 10, 0 < θ < 2 π

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

sin (x) = \frac{3}{5} sin (x + \frac{π}{2}) + cos (π - x)

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