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

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

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

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

Solution

cos (2 x) = 1 - sin (2 x), 0 \leq x \leq 2 π

Solution

x = 0, x = \frac{π}{4}, x = π, x = \frac{5 π}{4}, x = 2 π

Degrees

x = 0^{\circ}, x = 4 5^{\circ}, x = 18 0^{\circ}, x = 22 5^{\circ}, x = 36 0^{\circ}

Solution steps

cos (2 x) = 1 - sin (2 x), 0 \leq x \leq 2 π

Subtract

1 - sin (2 x)

from both sides

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

Rewrite using trig identities

cos (2 x) - 1 + sin (2 x)

sin (2 x) + cos (2 x) = 2 sin (2 x + \frac{π}{4})

sin (2 x) + cos (2 x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 x) + sin (\frac{π}{4}) cos (2 x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 x + \frac{π}{4})

= - 1 + 2 sin (2 x + \frac{π}{4})

- 1 + 2 sin (2 x + \frac{π}{4}) = 0

Move

1

to the right side

- 1 + 2 sin (2 x + \frac{π}{4}) = 0

Add

1

to both sides

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

Simplify

2 sin (2 x + \frac{π}{4}) = 1

2 sin (2 x + \frac{π}{4}) = 1

Divide both sides by

2

2 sin (2 x + \frac{π}{4}) = 1

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 sin ( 2 x + \frac{π}{4} )}{2} : sin (2 x + \frac{π}{4})

\frac{2 sin ( 2 x + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (2 x + \frac{π}{4})

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{π}{4}) = \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{2}

General solutions for

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

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

Solve

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

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

Subtract

\frac{π}{4}

from both sides

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

Simplify

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

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

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

Move

\frac{π}{4}

to the right side

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

Subtract

\frac{π}{4}

from both sides

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

Simplify

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

Simplify

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

Simplify

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

\frac{3 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

Apply rule

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

= \frac{- π + 3 π}{4}

Add similar elements:

- π + 3 π = 2 π

= \frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

Divide both sides by

2

2 x = 2 πn + \frac{π}{2}

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\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}

= πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn, x = πn + \frac{π}{4}

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = \frac{π}{4}, x = π, x = \frac{5 π}{4}, x = 2 π

Popular Examples

0=1+2sin(θ)

0 = 1 + 2 sin (θ)

csc(x-pi/3)=2

csc (x - \frac{π}{3}) = 2

sin(θ)=(sqrt(11))/6

sin (θ) = \frac{11}{6}

solvefor x,5cos(5x)=0 solve for

x, 5 cos (5 x) = 0

sin(3x)-sin(5x)=0

sin (3 x) - sin (5 x) = 0

Frequently Asked Questions (FAQ)

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