What is the general solution for sin(2x-1)=(sqrt(3))/2 ?

The general solution for sin(2x-1)=(sqrt(3))/2 is x=pin+1/2+pi/6 ,x=pin+1/2+pi/3

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sin(2x-1)=(sqrt(3))/2

Solution

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

Solution

x = πn + \frac{1}{2} + \frac{π}{6}, x = πn + \frac{1}{2} + \frac{π}{3}

Degrees

x = 58.64788 \dots^{\circ} + 18 0^{\circ} n, x = 88.64788 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

General solutions for

sin (2 x - 1) = \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 - 1 = \frac{π}{3} + 2 πn, 2 x - 1 = \frac{2 π}{3} + 2 πn

2 x - 1 = \frac{π}{3} + 2 πn, 2 x - 1 = \frac{2 π}{3} + 2 πn

Solve

2 x - 1 = \frac{π}{3} + 2 πn : x = πn + \frac{1}{2} + \frac{π}{6}

2 x - 1 = \frac{π}{3} + 2 πn

Move

1

to the right side

2 x - 1 = \frac{π}{3} + 2 πn

Add

1

to both sides

2 x - 1 + 1 = \frac{π}{3} + 2 πn + 1

Simplify

2 x = \frac{π}{3} + 2 πn + 1

2 x = \frac{π}{3} + 2 πn + 1

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn + 1

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{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{1}{2} : πn + \frac{1}{2} + \frac{π}{6}

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{2}

Group like terms

= \frac{1}{2} + \frac{2 πn}{2} + \frac{\frac{π}{3}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\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{1}{2} + πn + \frac{π}{6}

Group like terms

= πn + \frac{1}{2} + \frac{π}{6}

x = πn + \frac{1}{2} + \frac{π}{6}

x = πn + \frac{1}{2} + \frac{π}{6}

x = πn + \frac{1}{2} + \frac{π}{6}

Solve

2 x - 1 = \frac{2 π}{3} + 2 πn : x = πn + \frac{1}{2} + \frac{π}{3}

2 x - 1 = \frac{2 π}{3} + 2 πn

Move

1

to the right side

2 x - 1 = \frac{2 π}{3} + 2 πn

Add

1

to both sides

2 x - 1 + 1 = \frac{2 π}{3} + 2 πn + 1

Simplify

2 x = \frac{2 π}{3} + 2 πn + 1

2 x = \frac{2 π}{3} + 2 πn + 1

Divide both sides by

2

2 x = \frac{2 π}{3} + 2 πn + 1

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{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{1}{2} : πn + \frac{1}{2} + \frac{π}{3}

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2} + \frac{1}{2}

Group like terms

= \frac{1}{2} + \frac{2 πn}{2} + \frac{\frac{2 π}{3}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\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{1}{2} + πn + \frac{π}{3}

Group like terms

= πn + \frac{1}{2} + \frac{π}{3}

x = πn + \frac{1}{2} + \frac{π}{3}

x = πn + \frac{1}{2} + \frac{π}{3}

x = πn + \frac{1}{2} + \frac{π}{3}

x = πn + \frac{1}{2} + \frac{π}{6}, x = πn + \frac{1}{2} + \frac{π}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2x-1)=(sqrt(3))/2 ?
The general solution for sin(2x-1)=(sqrt(3))/2 is x=pin+1/2+pi/6 ,x=pin+1/2+pi/3