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

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

The general solution for sqrt(2)sin(2x)=1,0<= x<= 2pi is x= pi/8 ,x=(3pi)/8 ,x=(9pi)/8 ,x=(11pi)/8

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

Solution

2 sin (2 x) = 1, 0 \leq x \leq 2 π

Solution

x = \frac{π}{8}, x = \frac{3 π}{8}, x = \frac{9 π}{8}, x = \frac{11 π}{8}

Degrees

x = 22. 5^{\circ}, x = 67. 5^{\circ}, x = 202. 5^{\circ}, x = 247. 5^{\circ}

Solution steps

2 sin (2 x) = 1, 0 \leq x \leq 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

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

Simplify

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

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

Cancel the common factor:

2

= sin (2 x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

General solutions for

sin (2 x) = \frac{2}{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}{4} + 2 πn, 2 x = \frac{3 π}{4} + 2 πn

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

Solve

2 x = \frac{π}{4} + 2 πn : x = \frac{π}{8} + πn

2 x = \frac{π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{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}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

Solve

2 x = \frac{3 π}{4} + 2 πn : x = \frac{3 π}{8} + πn

2 x = \frac{3 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{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{3 π}{4}}{2} + \frac{2 πn}{2} : \frac{3 π}{8} + πn

\frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{π}{8} + πn, x = \frac{3 π}{8} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{8}, x = \frac{3 π}{8}, x = \frac{9 π}{8}, x = \frac{11 π}{8}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(2x)=1,0<= x<= 2pi ?
The general solution for sqrt(2)sin(2x)=1,0<= x<= 2pi is x= pi/8 ,x=(3pi)/8 ,x=(9pi)/8 ,x=(11pi)/8