What is the general solution for 2sin(4x)-1=0,0<x<= 2 ?

Q: What is the general solution for 2sin(4x)-1=0,0<x<= 2 ?

The general solution for 2sin(4x)-1=0,0<x<= 2 is x= pi/(24),x=(5pi)/(24),x=(13pi)/(24)

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2sin(4x)-1=0,0<x<= 2

Solution

2 sin (4 x) - 1 = 0, 0 < x \leq 2

Solution

x = \frac{π}{24}, x = \frac{5 π}{24}, x = \frac{13 π}{24}

Degrees

x = 7. 5^{\circ}, x = 37. 5^{\circ}, x = 97. 5^{\circ}

Solution steps

2 sin (4 x) - 1 = 0, 0 < x \leq 2

Move

1

to the right side

2 sin (4 x) - 1 = 0

Add

1

to both sides

2 sin (4 x) - 1 + 1 = 0 + 1

Simplify

2 sin (4 x) = 1

2 sin (4 x) = 1

Divide both sides by

2

2 sin (4 x) = 1

Divide both sides by

2

\frac{2 sin ( 4 x )}{2} = \frac{1}{2}

Simplify

sin (4 x) = \frac{1}{2}

sin (4 x) = \frac{1}{2}

General solutions for

sin (4 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}

4 x = \frac{π}{6} + 2 πn, 4 x = \frac{5 π}{6} + 2 πn

4 x = \frac{π}{6} + 2 πn, 4 x = \frac{5 π}{6} + 2 πn

Solve

4 x = \frac{π}{6} + 2 πn : x = \frac{π}{24} + \frac{πn}{2}

4 x = \frac{π}{6} + 2 πn

Divide both sides by

4

4 x = \frac{π}{6} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π}{24} + \frac{πn}{2}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{24} + \frac{πn}{2}

x = \frac{π}{24} + \frac{πn}{2}

x = \frac{π}{24} + \frac{πn}{2}

x = \frac{π}{24} + \frac{πn}{2}

Solve

4 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{24} + \frac{πn}{2}

4 x = \frac{5 π}{6} + 2 πn

Divide both sides by

4

4 x = \frac{5 π}{6} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4} : \frac{5 π}{24} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{5 π}{6}}{4} = \frac{5 π}{24}

\frac{\frac{5 π}{6}}{4}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{5 π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{5 π}{24} + \frac{πn}{2}

x = \frac{5 π}{24} + \frac{πn}{2}

x = \frac{5 π}{24} + \frac{πn}{2}

x = \frac{5 π}{24} + \frac{πn}{2}

x = \frac{π}{24} + \frac{πn}{2}, x = \frac{5 π}{24} + \frac{πn}{2}

Solutions for the range

0 < x \leq 2

x = \frac{π}{24}, x = \frac{5 π}{24}, x = \frac{13 π}{24}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(4x)-1=0,0<x<= 2 ?
The general solution for 2sin(4x)-1=0,0<x<= 2 is x= pi/(24),x=(5pi)/(24),x=(13pi)/(24)