What is the general solution for 2sin(2x)=1,0<x<2pi ?

Q: What is the general solution for 2sin(2x)=1,0<x<2pi ?

The general solution for 2sin(2x)=1,0<x<2pi is x= pi/(12),x=(5pi)/(12),x=(13pi)/(12),x=(17pi)/(12)

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

Solution

2 sin (2 x) = 1, 0 < x < 2 π

Solution

x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

Degrees

x = 1 5^{\circ}, x = 7 5^{\circ}, x = 19 5^{\circ}, x = 25 5^{\circ}

Solution steps

2 sin (2 x) = 1, 0 < x < 2 π

Divide both sides by

2

2 sin (2 x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (2 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}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{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}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{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{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solutions for the range

0 < x < 2 π

x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

Popular Examples

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

sin(x)=cos(2x),0<= x<= 2pi

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

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

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

sin(θ)=-(sqrt(5))/5 ,(3pi)/2 <θ<2pi

sin (θ) = - \frac{5}{5}, \frac{3 π}{2} < θ < 2 π

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

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

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)=1,0<x<2pi ?
The general solution for 2sin(2x)=1,0<x<2pi is x= pi/(12),x=(5pi)/(12),x=(13pi)/(12),x=(17pi)/(12)