What is the general solution for -2sqrt(2)=-4sin(2θ) ?

The general solution for -2sqrt(2)=-4sin(2θ) is θ= pi/8+pin,θ=(3pi)/8+pin

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-2sqrt(2)=-4sin(2θ)

- 22 = - 4 sin (2 θ)

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

Solution steps

- 22 = - 4 sin (2 θ)

Switch sides

- 4 sin (2 θ) = - 22

Divide both sides by

- 4

sin (2 θ) = \frac{2}{2}

General solutions for

sin (2 θ) = \frac{2}{2}

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

Solve

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

Solve

2 θ = \frac{3 π}{4} + 2 πn : θ = \frac{3 π}{8} + πn

θ = \frac{π}{8} + πn, θ = \frac{3 π}{8} + πn

8 cos (x) = 4 cos^{2} (x) + 3

\frac{sin ^{2} ( x )}{2} + c = cos (x) c + c

- tan (x) - sec (x) = 0

\frac{26}{sin ( 8 8 ^{\circ} )} = \frac{21}{sin ( c )}

3 tan^{2} (3 x) = 1

What is the general solution for -2sqrt(2)=-4sin(2θ) ?
The general solution for -2sqrt(2)=-4sin(2θ) is θ= pi/8+pin,θ=(3pi)/8+pin