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

The general solution for 2sin(3x)+1=0 is x=(7pi}{18}+(2pin)/3 ,x=(11pi)/(18)+\frac{2pin)/3

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2sin(3x)+1=0

2 sin (3 x) + 1 = 0

x = \frac{7 π}{18} + \frac{2 πn}{3}, x = \frac{11 π}{18} + \frac{2 πn}{3}

Solution steps

2 sin (3 x) + 1 = 0

Move

1

to the right side

2 sin (3 x) = - 1

Divide both sides by

2

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

General solutions for

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

3 x = \frac{7 π}{6} + 2 πn, 3 x = \frac{11 π}{6} + 2 πn

Solve

3 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{18} + \frac{2 πn}{3}

Solve

3 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}, x = \frac{11 π}{18} + \frac{2 πn}{3}

3 sin (θ) = cos (θ), 0 \leq θ \leq 2 π

cos (2 θ) = - \frac{2}{2}

tan (\frac{θ}{2} + \frac{π}{3}) = - 1

sin (2 x) = 3 cos (2 x)

sec (x) = - 2, 0 \leq x \leq 2 π

What is the general solution for 2sin(3x)+1=0 ?
The general solution for 2sin(3x)+1=0 is x=(7pi}{18}+(2pin)/3 ,x=(11pi)/(18)+\frac{2pin)/3