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

Q: What is the general solution for 2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi ?

The general solution for 2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi is No Solution for θ\in\mathbb{R}

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Popular >

2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi

Solution

2 sin (\frac{θ}{3}) + 3 = 0, 0 \leq θ \leq 2 π

Solution

No Solution for θ \in R

Solution steps

2 sin (\frac{θ}{3}) + 3 = 0, 0 \leq θ \leq 2 π

Move

3

to the right side

2 sin (\frac{θ}{3}) + 3 = 0

Subtract

3

from both sides

2 sin (\frac{θ}{3}) + 3 - 3 = 0 - 3

Simplify

2 sin (\frac{θ}{3}) = - 3

2 sin (\frac{θ}{3}) = - 3

Divide both sides by

2

2 sin (\frac{θ}{3}) = - 3

Divide both sides by

2

\frac{2 sin ( \frac{θ}{3} )}{2} = \frac{- 3}{2}

Simplify

sin (\frac{θ}{3}) = - \frac{3}{2}

sin (\frac{θ}{3}) = - \frac{3}{2}

General solutions for

sin (\frac{θ}{3}) = - \frac{3}{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}

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

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

Solve

\frac{θ}{3} = \frac{4 π}{3} + 2 πn : θ = 4 π + 6 πn

\frac{θ}{3} = \frac{4 π}{3} + 2 πn

Multiply both sides by

3

\frac{θ}{3} = \frac{4 π}{3} + 2 πn

Multiply both sides by

3

\frac{3 θ}{3} = 3 \cdot \frac{4 π}{3} + 3 \cdot 2 πn

Simplify

\frac{3 θ}{3} = 3 \cdot \frac{4 π}{3} + 3 \cdot 2 πn

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

3 \cdot \frac{4 π}{3} + 3 \cdot 2 πn : 4 π + 6 πn

3 \cdot \frac{4 π}{3} + 3 \cdot 2 πn

3 \cdot \frac{4 π}{3} = 4 π

3 \cdot \frac{4 π}{3}

Multiply fractions:

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

= \frac{4 π 3}{3}

Cancel the common factor:

3

= 4 π

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn

= 4 π + 6 πn

θ = 4 π + 6 πn

θ = 4 π + 6 πn

θ = 4 π + 6 πn

Solve

\frac{θ}{3} = \frac{5 π}{3} + 2 πn : θ = 5 π + 6 πn

\frac{θ}{3} = \frac{5 π}{3} + 2 πn

Multiply both sides by

3

\frac{θ}{3} = \frac{5 π}{3} + 2 πn

Multiply both sides by

3

\frac{3 θ}{3} = 3 \cdot \frac{5 π}{3} + 3 \cdot 2 πn

Simplify

\frac{3 θ}{3} = 3 \cdot \frac{5 π}{3} + 3 \cdot 2 πn

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

3 \cdot \frac{5 π}{3} + 3 \cdot 2 πn : 5 π + 6 πn

3 \cdot \frac{5 π}{3} + 3 \cdot 2 πn

3 \cdot \frac{5 π}{3} = 5 π

3 \cdot \frac{5 π}{3}

Multiply fractions:

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

= \frac{5 π 3}{3}

Cancel the common factor:

3

= 5 π

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn

= 5 π + 6 πn

θ = 5 π + 6 πn

θ = 5 π + 6 πn

θ = 5 π + 6 πn

θ = 4 π + 6 πn, θ = 5 π + 6 πn

Solutions for the range

0 \leq θ \leq 2 π

No Solution for θ \in R

Graph

Popular Examples

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

sin (2 x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

2sin(x-pi/3)+sqrt(3)=0

2 sin (x - \frac{π}{3}) + 3 = 0

sin(x)=0.77

sin (x) = 0.77

sin(x)=0.87

sin (x) = 0.87

sin(x)=0.14

sin (x) = 0.14

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi ?
The general solution for 2sin(θ/3)+sqrt(3)=0,0<= θ<= 2pi is No Solution for θ\in\mathbb{R}