What is the general solution for sqrt(2)sin(3θ)-1=0 ?

The general solution for sqrt(2)sin(3θ)-1=0 is θ= pi/(12)+(2pin)/3 ,θ= pi/4+(2pin)/3

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sqrt(2)sin(3θ)-1=0

2 sin (3 θ) - 1 = 0

θ = \frac{π}{12} + \frac{2 πn}{3}, θ = \frac{π}{4} + \frac{2 πn}{3}

Solution steps

2 sin (3 θ) - 1 = 0

Move

1

to the right side

2 sin (3 θ) = 1

Divide both sides by

2

sin (3 θ) = \frac{2}{2}

General solutions for

sin (3 θ) = \frac{2}{2}

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

Solve

3 θ = \frac{π}{4} + 2 πn : θ = \frac{π}{12} + \frac{2 πn}{3}

Solve

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

θ = \frac{π}{12} + \frac{2 πn}{3}, θ = \frac{π}{4} + \frac{2 πn}{3}

cot^{2} (θ) + 1 = \frac{3}{4}

- 2 cos (x) - cos (2 x) \cdot 2 = 0

tan^{2} (x) + 4 tan (x) + 2 = 0

- 9 sin^{2} (θ) + 4 = 5 cos (θ) - 1

tan (θ) = 1.25

What is the general solution for sqrt(2)sin(3θ)-1=0 ?
The general solution for sqrt(2)sin(3θ)-1=0 is θ= pi/(12)+(2pin)/3 ,θ= pi/4+(2pin)/3