What is the general solution for 3sin(x)=1+sin(x) ?

The general solution for 3sin(x)=1+sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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3sin(x)=1+sin(x)

3 sin (x) = 1 + sin (x)

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) = 1 + sin (x)

Solve by substitution

3 sin (x) = 1 + sin (x)

Let:

sin (x) = u

3 u = 1 + u

3 u = 1 + u : u = \frac{1}{2}

3 u = 1 + u

Move

u

to the left side

3 u = 1 + u

Subtract

u

from both sides

3 u - u = 1 + u - u

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

2 sin (x - \frac{π}{6}) + 1 = 0

3 - 3 sin (θ) = 6 cos^{2} (θ)

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

tan (3 x) = 5 tan (x)

2 cos (2 x) - 2 = 0

What is the general solution for 3sin(x)=1+sin(x) ?
The general solution for 3sin(x)=1+sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin