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

The general solution for sqrt(2)sin(2θ)+1=0 is θ=(5pi)/8+pin,θ=(7pi)/8+pin

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sqrt(2)sin(2θ)+1=0

2 sin (2 θ) + 1 = 0

θ = \frac{5 π}{8} + πn, θ = \frac{7 π}{8} + 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{2}{2}

General solutions for

sin (2 θ) = - \frac{2}{2}

2 θ = \frac{5 π}{4} + 2 πn, 2 θ = \frac{7 π}{4} + 2 πn

Solve

2 θ = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{8} + πn

Solve

2 θ = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{8} + πn

θ = \frac{5 π}{8} + πn, θ = \frac{7 π}{8} + πn

sin (2 x) - 2 sin (x) = 0

7 cos^{2} (x) - 1 = 5 cos (x)

x, sin (2 x) = 0

cot (θ) = - 4

- 4 sin (θ) + 5 = 3 sin (θ) + 5

What is the general solution for sqrt(2)sin(2θ)+1=0 ?
The general solution for sqrt(2)sin(2θ)+1=0 is θ=(5pi)/8+pin,θ=(7pi)/8+pin