What is the general solution for 3sin(2θ)=2sin(θ) ?

The general solution for 3sin(2θ)=2sin(θ) is θ=2pin,θ=pi+2pin,θ=1.23095…+2pin,θ=2pi-1.23095…+2pin

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Popular >

3sin(2θ)=2sin(θ)

Solution

3 sin (2 θ) = 2 sin (θ)

Solution

θ = 2 πn, θ = π + 2 πn, θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (2 θ) = 2 sin (θ)

Subtract

2 sin (θ)

from both sides

3 sin (2 θ) - 2 sin (θ) = 0

Rewrite using trig identities

- 2 sin (θ) + 3 sin (2 θ)

Use the Double Angle identity:

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

= - 2 sin (θ) + 3 \cdot 2 sin (θ) cos (θ)

Simplify

= - 2 sin (θ) + 6 sin (θ) cos (θ)

- 2 sin (θ) + 6 cos (θ) sin (θ) = 0

Factor

- 2 sin (θ) + 6 cos (θ) sin (θ) : 2 sin (θ) (3 cos (θ) - 1)

- 2 sin (θ) + 6 cos (θ) sin (θ)

Rewrite

6

3 \cdot 2

= - 2 sin (θ) + 3 \cdot 2 sin (θ) cos (θ)

Factor out common term

2 sin (θ)

= 2 sin (θ) (- 1 + 3 cos (θ))

2 sin (θ) (3 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 3 cos (θ) - 1 = 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

3 cos (θ) - 1 = 0 : θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

3 cos (θ) - 1 = 0

Move

1

to the right side

3 cos (θ) - 1 = 0

Add

1

to both sides

3 cos (θ) - 1 + 1 = 0 + 1

Simplify

3 cos (θ) = 1

3 cos (θ) = 1

Divide both sides by

3

3 cos (θ) = 1

Divide both sides by

3

\frac{3 cos ( θ )}{3} = \frac{1}{3}

Simplify

cos (θ) = \frac{1}{3}

cos (θ) = \frac{1}{3}

Apply trig inverse properties

cos (θ) = \frac{1}{3}

General solutions for

cos (θ) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(2θ)=2sin(θ) ?
The general solution for 3sin(2θ)=2sin(θ) is θ=2pin,θ=pi+2pin,θ=1.23095…+2pin,θ=2pi-1.23095…+2pin