What is the general solution for 2sin(2θ)-sqrt(3)=0 ?

The general solution for 2sin(2θ)-sqrt(3)=0 is θ= pi/6+pin,θ= pi/3+pin

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2sin(2θ)-sqrt(3)=0

2 sin (2 θ) - 3 = 0

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

Solution steps

2 sin (2 θ) - 3 = 0

Move

3

to the right side

2 sin (2 θ) = 3

Divide both sides by

2

sin (2 θ) = \frac{3}{2}

General solutions for

sin (2 θ) = \frac{3}{2}

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

Solve

2 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{6} + πn

Solve

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

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

2 sin^{2} (x) - sin (x) - 1 = 0, 0 \leq x \leq 2 π

4 cos (2 θ) + 12 = - 2 cos (θ) + 9

sin^{2} (2 θ) = 1

sin (x) = \frac{6}{7}

x, sin (x) = - \frac{1}{2}

What is the general solution for 2sin(2θ)-sqrt(3)=0 ?
The general solution for 2sin(2θ)-sqrt(3)=0 is θ= pi/6+pin,θ= pi/3+pin