What is the general solution for 4sin(θ)=sin(2θ) ?

The general solution for 4sin(θ)=sin(2θ) is θ=2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4sin(θ)=sin(2θ)

4 sin (θ) = sin (2 θ)

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) = sin (2 θ)

Subtract

sin (2 θ)

from both sides

4 sin (θ) - sin (2 θ) = 0

Rewrite using trig identities

- sin (2 θ) + 4 sin (θ)

Use the Double Angle identity:

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

= - 2 sin (θ) cos (θ) + 4 sin (θ)

4 sin (θ) - 2 cos (θ) sin (θ) = 0

Factor

4 sin (θ) - 2 cos (θ) sin (θ) : - 2 sin (θ) (cos (θ) - 2)

4 sin (θ) - 2 cos (θ) sin (θ)

Rewrite

4

2 \cdot 2

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

Factor out common term

- 2 sin (θ)

= - 2 sin (θ) (- 2 + cos (θ))

- 2 sin (θ) (cos (θ) - 2) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) - 2 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (θ) - 2 = 0 :

No Solution

cos (θ) - 2 = 0

Move

2

to the right side

cos (θ) - 2 = 0

Add

2

to both sides

cos (θ) - 2 + 2 = 0 + 2

Simplify

cos (θ) = 2

cos (θ) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

What is the general solution for 4sin(θ)=sin(2θ) ?
The general solution for 4sin(θ)=sin(2θ) is θ=2pin,θ=pi+2pin