What is the general solution for sin(x)=sqrt(3)-sin(x) ?

The general solution for sin(x)=sqrt(3)-sin(x) is x= pi/3+2pin,x=(2pi)/3+2pin

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sin(x)=sqrt(3)-sin(x)

sin (x) = 3 - sin (x)

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) = 3 - sin (x)

Solve by substitution

sin (x) = 3 - sin (x)

Let:

sin (x) = u

u = 3 - u

u = 3 - u : u = \frac{3}{2}

u = 3 - u

Move

u

to the left side

u = 3 - u

Add

u

to both sides

u + u = 3 - u + u

Simplify

2 u = 3

2 u = 3

Divide both sides by

2

2 u = 3

Divide both sides by

2

\frac{2 u}{2} = \frac{3}{2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = \frac{3}{2}

sin (x) = \frac{3}{2}

sin (x) = \frac{3}{2} : x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

sin (x) = \frac{3}{2}

General solutions for

sin (x) = \frac{3}{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}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

cot (θ) = - \frac{3}{3}

sin (x) - tan (x) = 0

sin (x + 2 0^{\circ}) = cos (x)

cos (x) - 3 sin (x) = 1

1 - sin (x) = cos (x)

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