What is the general solution for 2sin(x)=sqrt(2)sin(2x) ?

The general solution for 2sin(x)=sqrt(2)sin(2x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(7pi)/4+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(x)=sqrt(2)sin(2x)

Solution

2 sin (x) = 2 sin (2 x)

Solution

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) = 2 sin (2 x)

Subtract

2 sin (2 x)

from both sides

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

Rewrite using trig identities

2 sin (x) - sin (2 x) 2

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (x) - 2 \cdot 2 sin (x) cos (x)

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

Factor

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

2 sin (x) - 2 cos (x) sin (x) 2

Rewrite as

= 1 \cdot 2 sin (x) - 2 sin (x) cos (x) 2

Factor out common term

2 sin (x)

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

1 - 2 cos (x) = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

1 - 2 cos (x) = 0

Move

1

to the right side

1 - 2 cos (x) = 0

Subtract

1

from both sides

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

Simplify

- 2 cos (x) = - 1

- 2 cos (x) = - 1

Divide both sides by

- 2

- 2 cos (x) = - 1

Divide both sides by

- 2

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

Simplify

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

Simplify

\frac{- 2 cos ( x )}{- 2} : cos (x)

\frac{- 2 cos ( x )}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

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

Cancel the common factor:

2

= cos (x)

Simplify

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

\frac{- 1}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{1}{2}

Rationalize

\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}

= \frac{2}{2}

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

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

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Popular Examples

sin(θ)+cos(θ)= 7/5 ,sin(2θ)

sin (θ) + cos (θ) = \frac{7}{5}, sin (2 θ)

tan^2(x)+tan(x)=2

tan^{2} (x) + tan (x) = 2

solvefor θ,r=4cos(θ)solve for

θ, r = 4 cos (θ)

tan(x)= 5/9

tan (x) = \frac{5}{9}

sin^2(x)-sin(x)=5cos^2(x)-4

sin^{2} (x) - sin (x) = 5 cos^{2} (x) - 4

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)=sqrt(2)sin(2x) ?
The general solution for 2sin(x)=sqrt(2)sin(2x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(7pi)/4+2pin