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

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

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1-sin(2x)=0

Solution

1 - sin (2 x) = 0

Solution

x = \frac{π}{4} + πn

Degrees

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

Solution steps

1 - sin (2 x) = 0

Move

1

to the right side

1 - sin (2 x) = 0

Subtract

1

from both sides

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

Simplify

- sin (2 x) = - 1

- sin (2 x) = - 1

Divide both sides by

- 1

- sin (2 x) = - 1

Divide both sides by

- 1

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

Popular Examples

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2 cos (3 x) + cos (2 x) + 1 = 0

4+2tan^2(θ)=5tan^2(θ)+3

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sin(x)=sin(pi/5)

sin (x) = sin (\frac{π}{5})

Frequently Asked Questions (FAQ)

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