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

The general solution for 3sin(x)+1=sin(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

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

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

Degrees

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

1

to the right side

3 u + 1 = u

Subtract

1

from both sides

3 u + 1 - 1 = u - 1

Simplify

3 u = u - 1

3 u = u - 1

Move

u

to the left side

3 u = u - 1

Subtract

u

from both sides

3 u - u = u - 1 - 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{7 π}{6} + 2 πn, x = \frac{11 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

Combine all the solutions

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

cos (2 θ - \frac{π}{2}) = 1

- sin (2 x) + cos (x) = 0

3 csc^{2} (x) = 4

6 sec^{2} (x) + 3 tan^{2} (x) - 9 = 0

(3 tan^{2} (x) - 1) (tan^{2} (x) - 3) = 0

What is the general solution for 3sin(x)+1=sin(x) ?
The general solution for 3sin(x)+1=sin(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin