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 \leq 2 π

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

Solution steps

sin (3 θ) = - 1, 0 \leq θ \leq 2 π

General solutions for

sin (3 θ) = - 1

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}

θ = \frac{π}{2} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

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

cos (x - \frac{π}{3}) = \frac{1}{2}

tan (x) = \frac{7}{5}

- 2 cos (2 x) = 0

sin (7 x) - 1 = 0

cos (x) (2 sin (x) - 1) = 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