What is the general solution for sin(3θ)=-1,0<= θ<2pi ?

The general solution for sin(3θ)=-1,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6

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sin(3θ)=-1,0<= θ<2pi

sin (3 θ) = - 1, 0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 9 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

sin (3 θ) = - 1, 0 \leq θ < 2 π

General solutions for

sin (3 θ) = - 1

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 = \frac{3 π}{2} + 2 πn

3 θ = \frac{3 π}{2} + 2 πn

Solve

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

3 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

cos (θ) = cos (arctan (\frac{3}{4}))

13 sin (2 x) - 3 = 3

sin (x) = \frac{4}{9}

sin (x) = \frac{4}{6}

- 4 sin (x) = 0

What is the general solution for sin(3θ)=-1,0<= θ<2pi ?
The general solution for sin(3θ)=-1,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6