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

The general solution for sqrt(2)sin(3θ)+1=0 is θ=(5pi}{12}+(2pin)/3 ,θ=(7pi)/(12)+\frac{2pin)/3

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sqrt(2)sin(3θ)+1=0

2 sin (3 θ) + 1 = 0

θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \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{5 π}{4} + 2 πn, 3 θ = \frac{7 π}{4} + 2 πn

Solve

3 θ = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{12} + \frac{2 πn}{3}

Solve

3 θ = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{12} + \frac{2 πn}{3}

θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \frac{2 πn}{3}

sin (a) = 0.7245

3 tan (\frac{x}{4}) + 5 = 4

2 sin (3 x) = 1

sin (4 x) = cos (2 x)

cot^{2} (θ) - cot (θ) = 0

What is the general solution for sqrt(2)sin(3θ)+1=0 ?
The general solution for sqrt(2)sin(3θ)+1=0 is θ=(5pi}{12}+(2pin)/3 ,θ=(7pi)/(12)+\frac{2pin)/3