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

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

The general solution for sqrt(2)sin(2x)=sin(4x),0<= x<= pi is x=0,x= pi/2 ,x=pi,x= pi/8 ,x=(7pi)/8

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Popular Trigonometry >

sqrt(2)sin(2x)=sin(4x),0<= x<= pi

Solution

2 sin (2 x) = sin (4 x), 0 \leq x \leq π

Solution

x = 0, x = \frac{π}{2}, x = π, x = \frac{π}{8}, x = \frac{7 π}{8}

Degrees

x = 0^{\circ}, x = 9 0^{\circ}, x = 18 0^{\circ}, x = 22. 5^{\circ}, x = 157. 5^{\circ}

Solution steps

2 sin (2 x) = sin (4 x), 0 \leq x \leq π

Subtract

sin (4 x)

from both sides

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

Let:

u = 2 x

2 sin (u) - sin (2 u) = 0, 0 \leq u \leq 2 π

Rewrite using trig identities

- sin (2 u) + sin (u) 2

Use the Double Angle identity:

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

= - 2 sin (u) cos (u) + 2 sin (u)

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

Factor

sin (u) 2 - 2 cos (u) sin (u) : sin (u) (2 - 2 cos (u))

sin (u) 2 - 2 cos (u) sin (u)

Factor out common term

sin (u)

= sin (u) (2 - 2 cos (u))

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

Solving each part separately

sin (u) = 0 or 2 - 2 cos (u) = 0

sin (u) = 0, 0 \leq u \leq 2 π : u = 0, u = π, u = 2 π

sin (u) = 0, 0 \leq u \leq 2 π

General solutions for

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

Solutions for the range

0 \leq u \leq 2 π

u = 0, u = π, u = 2 π

2 - 2 cos (u) = 0, 0 \leq u \leq 2 π : u = \frac{π}{4}, u = \frac{7 π}{4}

2 - 2 cos (u) = 0, 0 \leq u \leq 2 π

Move

2

to the right side

2 - 2 cos (u) = 0

Subtract

2

from both sides

2 - 2 cos (u) - 2 = 0 - 2

Simplify

- 2 cos (u) = - 2

- 2 cos (u) = - 2

Divide both sides by

- 2

- 2 cos (u) = - 2

Divide both sides by

- 2

\frac{- 2 cos ( u )}{- 2} = \frac{- 2}{- 2}

Simplify

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

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

General solutions for

cos (u) = \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}

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

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

Solutions for the range

0 \leq u \leq 2 π

u = \frac{π}{4}, u = \frac{7 π}{4}

Combine all the solutions

u = 0, u = π, u = 2 π, u = \frac{π}{4}, u = \frac{7 π}{4}

Substitute back

u = 2 x

2 x = 0 : x = 0

2 x = 0

Divide both sides by

2

2 x = 0

Divide both sides by

2

\frac{2 x}{2} = \frac{0}{2}

Simplify

x = 0

x = 0

2 x = π : x = \frac{π}{2}

2 x = π

Divide both sides by

2

2 x = π

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2}

Simplify

x = \frac{π}{2}

x = \frac{π}{2}

2 x = 2 π : x = π

2 x = 2 π

Divide both sides by

2

2 x = 2 π

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2}

Simplify

x = π

x = π

2 x = \frac{π}{4} : x = \frac{π}{8}

2 x = \frac{π}{4}

Divide both sides by

2

2 x = \frac{π}{4}

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{4}}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{4}}{2} : \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

x = \frac{π}{8}

x = \frac{π}{8}

x = \frac{π}{8}

2 x = \frac{7 π}{4} : x = \frac{7 π}{8}

2 x = \frac{7 π}{4}

Divide both sides by

2

2 x = \frac{7 π}{4}

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{7 π}{4}}{2} : \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{7 π}{8}

x = \frac{7 π}{8}

x = \frac{7 π}{8}

x = \frac{7 π}{8}

x = 0, x = \frac{π}{2}, x = π, x = \frac{π}{8}, x = \frac{7 π}{8}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(2x)=sin(4x),0<= x<= pi ?
The general solution for sqrt(2)sin(2x)=sin(4x),0<= x<= pi is x=0,x= pi/2 ,x=pi,x= pi/8 ,x=(7pi)/8