What is the general solution for sin(θ)=5sin(θ)+2 ?

The general solution for sin(θ)=5sin(θ)+2 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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Popular >

sin(θ)=5sin(θ)+2

Solution

sin (θ) = 5 sin (θ) + 2

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = 5 sin (θ) + 2

Solve by substitution

sin (θ) = 5 sin (θ) + 2

Let:

sin (θ) = u

u = 5 u + 2

u = 5 u + 2 : u = - \frac{1}{2}

u = 5 u + 2

Move

5 u

to the left side

u = 5 u + 2

Subtract

5 u

from both sides

u - 5 u = 5 u + 2 - 5 u

Simplify

- 4 u = 2

- 4 u = 2

Divide both sides by

- 4

- 4 u = 2

Divide both sides by

- 4

\frac{- 4 u}{- 4} = \frac{2}{- 4}

Simplify

\frac{- 4 u}{- 4} = \frac{2}{- 4}

Simplify

\frac{- 4 u}{- 4} : u

\frac{- 4 u}{- 4}

Apply the fraction rule:

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

= \frac{4 u}{4}

Divide the numbers:

\frac{4}{4} = 1

= u

Simplify

\frac{2}{- 4} : - \frac{1}{2}

\frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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 = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Popular Examples

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

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

2sin(3x+1)=1

2 sin (3 x + 1) = 1

1-cos(θ)=0

1 - cos (θ) = 0

sin^2(x)+sin^2(2x)=1

sin^{2} (x) + sin^{2} (2 x) = 1

tan(2x-5)=1

tan (2 x - 5) = 1

Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)=5sin(θ)+2 ?
The general solution for sin(θ)=5sin(θ)+2 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin