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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Popular >

2sin(3x)+1=0

Solution

2 sin (3 x) + 1 = 0

Solution

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

Degrees

x = 7 0^{\circ} + 12 0^{\circ} n, x = 11 0^{\circ} + 12 0^{\circ} n

Solution steps

2 sin (3 x) + 1 = 0

Move

1

to the right side

2 sin (3 x) + 1 = 0

Subtract

1

from both sides

2 sin (3 x) + 1 - 1 = 0 - 1

Simplify

2 sin (3 x) = - 1

2 sin (3 x) = - 1

Divide both sides by

2

2 sin (3 x) = - 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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}

3 x = \frac{7 π}{6} + 2 πn

Divide both sides by

3

3 x = \frac{7 π}{6} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

\frac{\frac{7 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{7 π}{6}}{3} = \frac{7 π}{18}

\frac{\frac{7 π}{6}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{7 π}{18}

= \frac{7 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}

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}

3 x = \frac{11 π}{6} + 2 πn

Divide both sides by

3

3 x = \frac{11 π}{6} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

\frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{11 π}{6}}{3} = \frac{11 π}{18}

\frac{\frac{11 π}{6}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{11 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{11 π}{18}

= \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{11 π}{18} + \frac{2 πn}{3}

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}

Popular Examples

sqrt(3)sin(θ)=cos(θ),0<= θ<= 2pi

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

cos(2θ)=-(sqrt(2))/2

cos (2 θ) = - \frac{2}{2}

tan(θ/2+pi/3)=-1

tan (\frac{θ}{2} + \frac{π}{3}) = - 1

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

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

sec(x)=-sqrt(2),0<= x<= 2pi

sec (x) = - 2, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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