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

The general solution for (sin(2x)-1)(2sin(2x)-1)=0 is x= pi/4+pin,x= pi/(12)+pin,x=(5pi)/(12)+pin

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

Solution

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

Solution

x = \frac{π}{4} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solving each part separately

sin (2 x) - 1 = 0 or 2 sin (2 x) - 1 = 0

sin (2 x) - 1 = 0 : x = \frac{π}{4} + πn

sin (2 x) - 1 = 0

Move

1

to the right side

sin (2 x) - 1 = 0

Add

1

to both sides

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

Simplify

sin (2 x) = 1

sin (2 x) = 1

General solutions for

sin (2 x) = 1

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}{2} + 2 πn

2 x = \frac{π}{2} + 2 πn

Solve

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

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

2 sin (2 x) - 1 = 0 : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

2 sin (2 x) - 1 = 0

Move

1

to the right side

2 sin (2 x) - 1 = 0

Add

1

to both sides

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

Simplify

2 sin (2 x) = 1

2 sin (2 x) = 1

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

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x)-1)(2sin(2x)-1)=0 ?
The general solution for (sin(2x)-1)(2sin(2x)-1)=0 is x= pi/4+pin,x= pi/(12)+pin,x=(5pi)/(12)+pin