What is the general solution for 2sin(2θ)+1=0 ?

The general solution for 2sin(2θ)+1=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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2sin(2θ)+1=0

2 sin (2 θ) + 1 = 0

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

Solution steps

2 sin (2 θ) + 1 = 0

Move

1

to the right side

2 sin (2 θ) = - 1

Divide both sides by

2

sin (2 θ) = - \frac{1}{2}

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

2 sin (θ) - sin (2 θ) = 0

3 tan^{2} (x) + 4 tan (x) - 4 = 0

csc^{2} (θ) + 2 cot (θ) = 0

sin (x) cos (x) = - 10 cos (x)

tan^{2} (2 x) = 1

What is the general solution for 2sin(2θ)+1=0 ?
The general solution for 2sin(2θ)+1=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin