What is the general solution for sin(3θ)=-1,0<= θ<= 360 ?

The general solution for sin(3θ)=-1,0<= θ<= 360 is θ=90,θ=210,θ=330

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sin(3θ)=-1,0<= θ<= 360

sin (3 θ) = - 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

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

Radians

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

Solution steps

sin (3 θ) = - 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

General solutions for

sin (3 θ) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

3 θ = 27 0^{\circ} + 36 0^{\circ} n

3 θ = 27 0^{\circ} + 36 0^{\circ} n

Solve

3 θ = 27 0^{\circ} + 36 0^{\circ} n : θ = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

3 θ = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

3

3 θ = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

3

\frac{3 θ}{3} = \frac{27 0 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

\frac{3 θ}{3} = \frac{27 0 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{27 0 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} : 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

\frac{27 0 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

\frac{27 0 ^{\circ}}{3} = 9 0^{\circ}

\frac{27 0 ^{\circ}}{3}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= 9 0^{\circ}

Cancel the common factor:

3

= 9 0^{\circ}

= 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

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

What is the general solution for sin(3θ)=-1,0<= θ<= 360 ?
The general solution for sin(3θ)=-1,0<= θ<= 360 is θ=90,θ=210,θ=330