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

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

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2sqrt(3)sin(2x)-3=0

Solution

23 sin (2 x) - 3 = 0

Solution

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

Degrees

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

Solution steps

23 sin (2 x) - 3 = 0

Move

3

to the right side

23 sin (2 x) - 3 = 0

Add

3

to both sides

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

Simplify

23 sin (2 x) = 3

23 sin (2 x) = 3

Divide both sides by

23

23 sin (2 x) = 3

Divide both sides by

23

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

Simplify

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

Simplify

\frac{2 3 sin ( 2 x )}{2 3} : sin (2 x)

\frac{2 3 sin ( 2 x )}{2 3}

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

3

= sin (2 x)

Simplify

\frac{3}{2 3} : \frac{3}{2}

\frac{3}{2 3}

Apply radical rule:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3}{2 \cdot 3 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= \frac{3 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{3}{2}

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

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

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

General solutions for

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

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

Solve

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

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

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

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

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

Graph

Popular Examples

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sec (x) = \frac{3}{2}

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sin (2 x) = 3

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sin(x)+cos(x)=(sqrt(6))/2

sin (x) + cos (x) = \frac{6}{2}

Frequently Asked Questions (FAQ)

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