What is the general solution for 2sin(3x+(3pi)/2)=1 ?

The general solution for 2sin(3x+(3pi)/2)=1 is x=(2pin)/3-(4pi)/9 ,x=(2pin)/3-(2pi)/9

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2sin(3x+(3pi)/2)=1

2 sin (3 x + \frac{3 π}{2}) = 1

x = \frac{2 πn}{3} - \frac{4 π}{9}, x = \frac{2 πn}{3} - \frac{2 π}{9}

Solution steps

2 sin (3 x + \frac{3 π}{2}) = 1

Divide both sides by

2

sin (3 x + \frac{3 π}{2}) = \frac{1}{2}

General solutions for

sin (3 x + \frac{3 π}{2}) = \frac{1}{2}

3 x + \frac{3 π}{2} = \frac{π}{6} + 2 πn, 3 x + \frac{3 π}{2} = \frac{5 π}{6} + 2 πn

Solve

3 x + \frac{3 π}{2} = \frac{π}{6} + 2 πn : x = \frac{2 πn}{3} - \frac{4 π}{9}

Solve

3 x + \frac{3 π}{2} = \frac{5 π}{6} + 2 πn : x = \frac{2 πn}{3} - \frac{2 π}{9}

x = \frac{2 πn}{3} - \frac{4 π}{9}, x = \frac{2 πn}{3} - \frac{2 π}{9}

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What is the general solution for 2sin(3x+(3pi)/2)=1 ?
The general solution for 2sin(3x+(3pi)/2)=1 is x=(2pin)/3-(4pi)/9 ,x=(2pin)/3-(2pi)/9