What is the general solution for 2sin(2θ)=-sqrt(2) ?

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

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2sin(2θ)=-sqrt(2)

2 sin (2 θ) = - 2

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

Solution steps

2 sin (2 θ) = - 2

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

2 sin (2 x) = 2

2 sec (θ) + 1 = 0

3 arccos (2 x) = 2 π

3 cos (x) = 1

cos (a) = \frac{4}{5}

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