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

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

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2sin(θ)=-1,0<= θ<2pi

2 sin (θ) = - 1, 0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Solution steps

2 sin (θ) = - 1, 0 \leq θ < 2 π

Divide both sides by

2

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

cos (16 x) = 0

sin (θ) = - \frac{35}{6}

0.6 = cos (x)

3 tan^{3} (x) - 1 = 0

- 2 csc^{2} (2 x) = 0

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