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}

Solution steps

sin (3 θ) = 1, 0 \leq θ < 2 π

General solutions for

sin (3 θ) = 1

3 θ = \frac{π}{2} + 2 πn

Solve

3 θ = \frac{π}{2} + 2 πn : 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