What is the general solution for 4sin(θ)=1+sin(θ) ?

The general solution for 4sin(θ)=1+sin(θ) is θ=0.33983…+2pin,θ=pi-0.33983…+2pin

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4sin(θ)=1+sin(θ)

4 sin (θ) = 1 + sin (θ)

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

Degrees

θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) = 1 + sin (θ)

Solve by substitution

4 sin (θ) = 1 + sin (θ)

Let:

sin (θ) = u

4 u = 1 + u

4 u = 1 + u : u = \frac{1}{3}

4 u = 1 + u

Move

u

to the left side

4 u = 1 + u

Subtract

u

from both sides

4 u - u = 1 + u - u

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

\frac{3 u}{3} = \frac{1}{3}

Simplify

u = \frac{1}{3}

u = \frac{1}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{3}

sin (θ) = \frac{1}{3}

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

sin (θ) = \frac{1}{3}

Apply trig inverse properties

sin (θ) = \frac{1}{3}

General solutions for

sin (θ) = \frac{1}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

tan (θ) - 3 = 0, 0 \leq θ \leq 2 π

\frac{1}{32} \cdot 120 0^{2} \cdot sin (2 θ) = 3000

1 - cos^{2} (a) = sin^{2} (2 a)

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

tan^{2} (θ) + 2 tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0

What is the general solution for 4sin(θ)=1+sin(θ) ?
The general solution for 4sin(θ)=1+sin(θ) is θ=0.33983…+2pin,θ=pi-0.33983…+2pin