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

The general solution for sin(2θ)=-1/2 ,θ>= 0 is

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sin(2θ)=-1/2 ,θ>= 0

sin (2 θ) = - \frac{1}{2}, θ \geq 0

θ = \frac{7 π}{12}, θ = \frac{11 π}{12}, θ = \frac{19 π}{12}, θ = \frac{23 π}{12}, θ = \frac{31 π}{12}, θ = \frac{35 π}{12}, θ = \frac{43 π}{12}, θ = \frac{47 π}{12}, θ = \frac{55 π}{12}, θ = \frac{59 π}{12}

Solution steps

sin (2 θ) = - \frac{1}{2}, θ \geq 0

General solutions for

sin (2 θ) = - \frac{1}{2}

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

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Solutions for the range

θ \geq 0

θ = \frac{7 π}{12}, θ = \frac{11 π}{12}, θ = \frac{19 π}{12}, θ = \frac{23 π}{12}, θ = \frac{31 π}{12}, θ = \frac{35 π}{12}, θ = \frac{43 π}{12}, θ = \frac{47 π}{12}, θ = \frac{55 π}{12}, θ = \frac{59 π}{12}

tan (x) = \frac{1}{3}, 0 \leq x \leq 2 π

2 sin (x) + 2 = 3 cos^{2} (x)

4 sin (2 x) + 5 cos (2 x) = 0

sin (x) = cot (x)

2 cos^{2} (\frac{x}{2}) - 3 cos (x) = 0

What is the general solution for sin(2θ)=-1/2 ,θ>= 0 ?
The general solution for sin(2θ)=-1/2 ,θ>= 0 is