What is the general solution for 1=1+3sin(3θ) ?

The general solution for 1=1+3sin(3θ) is θ=(2pin)/3 ,θ= pi/3+(2pin)/3

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1=1+3sin(3θ)

1 = 1 + 3 sin (3 θ)

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

Degrees

θ = 0^{\circ} + 12 0^{\circ} n, θ = 6 0^{\circ} + 12 0^{\circ} n

Solution steps

1 = 1 + 3 sin (3 θ)

Switch sides

1 + 3 sin (3 θ) = 1

Subtract

1

from both sides

1 + 3 sin (3 θ) - 1 = 1 - 1

Simplify

3 sin (3 θ) = 0

Divide both sides by

3

3 sin (3 θ) = 0

Divide both sides by

3

\frac{3 sin ( 3 θ )}{3} = \frac{0}{3}

Simplify

sin (3 θ) = 0

sin (3 θ) = 0

General solutions for

sin (3 θ) = 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}

3 θ = 0 + 2 πn, 3 θ = π + 2 πn

3 θ = 0 + 2 πn, 3 θ = π + 2 πn

Solve

3 θ = 0 + 2 πn : θ = \frac{2 πn}{3}

3 θ = 0 + 2 πn

0 + 2 πn = 2 πn

3 θ = 2 πn

Divide both sides by

3

3 θ = 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{2 πn}{3}

Simplify

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

Solve

3 θ = π + 2 πn : θ = \frac{π}{3} + \frac{2 πn}{3}

3 θ = π + 2 πn

Divide both sides by

3

3 θ = π + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{π}{3} + \frac{2 πn}{3}

Simplify

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

cos (\frac{θ}{4}) = \frac{2}{2}

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

cosh (x) + 4 sinh (x) + 2 = 0

cos (b) = \frac{5}{13}

x, sin (x) = \frac{1}{3}

What is the general solution for 1=1+3sin(3θ) ?
The general solution for 1=1+3sin(3θ) is θ=(2pin)/3 ,θ= pi/3+(2pin)/3