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

The general solution for sin(2θ)+1=0 is θ=(3pi)/4+pin

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sin(2θ)+1=0

sin (2 θ) + 1 = 0

θ = \frac{3 π}{4} + πn

Solution steps

sin (2 θ) + 1 = 0

Move

1

to the right side

sin (2 θ) = - 1

General solutions for

sin (2 θ) = - 1

2 θ = \frac{3 π}{2} + 2 πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

x, cos (x) = cos (y)

3 tanh^{2} (θ) = 5 sech (θ) + 1

6 cos (x) + 6 sin (x) tan (x) = 12

cos (θ) = \frac{8}{17}, 27 0^{\circ} < θ < 36 0^{\circ}

2 csc (θ) - 3 = 0

What is the general solution for sin(2θ)+1=0 ?
The general solution for sin(2θ)+1=0 is θ=(3pi)/4+pin