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/6 ,θ=(5pi)/6 ,θ=(3pi)/2

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sin(3θ)=1,0<= θ<2pi

sin (3 θ) = 1, 0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{3 π}{2}

Degrees

θ = 3 0^{\circ}, θ = 15 0^{\circ}, θ = 27 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{π}{2} + 2 πn

3 θ = \frac{π}{2} + 2 πn

Solve

3 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{6} + \frac{2 πn}{3}

3 θ = \frac{π}{2} + 2 πn

Divide both sides by

3

3 θ = \frac{π}{2} + 2 πn

Divide both sides by

3

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

Simplify

\frac{3 θ}{3} = \frac{\frac{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{π}{2}}{3} + \frac{2 πn}{3} : \frac{π}{6} + \frac{2 πn}{3}

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

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{3 π}{2}

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

sin (2 x) cos (x) = sin (x)

tan (\frac{1}{2} x) - 1 = 0

sin (3 x) = 0, 0 \leq x \leq 2 π

cos (x) - 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/6 ,θ=(5pi)/6 ,θ=(3pi)/2