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

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

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2sin(3θ)+sqrt(2)=0

2 sin (3 θ) + 2 = 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) + 2 = 0

Move

2

to the right side

2 sin (3 θ) = - 2

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}

sec (θ) = - \frac{13}{12}

0 = - sin (x) - cos (x)

9 sin (θ) - 7 = 0

csc (2 x) = 2

sin (4 x) cos (x) - sin (x) cos (4 x) = \frac{3}{2}

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