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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

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 (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)) = 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

2 - 2 cos (u) = 0, 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 = π : x = \frac{π}{2}

2 x = 2 π : x = π

2 x = \frac{π}{4} : x = \frac{π}{8}

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

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

9 sin (x) - 2 = 4 sin^{2} (x)

- a c_{1} sin (a x) + a c_{2} cos (a x) = 0

3 - 3 cos (x) = 0

- 2 cot (x) = csc^{2} (x)

sin (θ) = - 0.671

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