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

The general solution for 2sin(2θ)-sqrt(3)=0 is θ= pi/6+pin,θ= pi/3+pin

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2sin(2θ)-sqrt(3)=0

Solution

2 sin (2 θ) - 3 = 0

Solution

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (2 θ) - 3 = 0

Move

3

to the right side

2 sin (2 θ) - 3 = 0

Add

3

to both sides

2 sin (2 θ) - 3 + 3 = 0 + 3

Simplify

2 sin (2 θ) = 3

2 sin (2 θ) = 3

Divide both sides by

2

2 sin (2 θ) = 3

Divide both sides by

2

\frac{2 sin ( 2 θ )}{2} = \frac{3}{2}

Simplify

sin (2 θ) = \frac{3}{2}

sin (2 θ) = \frac{3}{2}

General solutions for

sin (2 θ) = \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}

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

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

Solve

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

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Solve

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

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{3} + πn

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

Popular Examples

2sin^2(x)-sin(x)-1=0,0<= x<= 2pi

2 sin^{2} (x) - sin (x) - 1 = 0, 0 \leq x \leq 2 π

4cos(2θ)+12=-2cos(θ)+9

4 cos (2 θ) + 12 = - 2 cos (θ) + 9

sin^2(2θ)=1

sin^{2} (2 θ) = 1

sin(x)= 6/7

sin (x) = \frac{6}{7}

solvefor x,sin(x)=-1/2 solve for

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

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2θ)-sqrt(3)=0 ?
The general solution for 2sin(2θ)-sqrt(3)=0 is θ= pi/6+pin,θ= pi/3+pin