What is the general solution for 3sin(x)+sqrt(2)=sin(x) ?

The general solution for 3sin(x)+sqrt(2)=sin(x) is x=(5pi)/4+2pin,x=(7pi)/4+2pin

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3sin(x)+sqrt(2)=sin(x)

3 sin (x) + 2 = sin (x)

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) + 2 = sin (x)

Solve by substitution

3 sin (x) + 2 = sin (x)

Let:

sin (x) = u

3 u + 2 = u

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

3 u + 2 = u

Move

2

to the right side

3 u + 2 = u

Subtract

2

from both sides

3 u + 2 - 2 = u - 2

Simplify

3 u = u - 2

3 u = u - 2

Move

u

to the left side

3 u = u - 2

Subtract

u

from both sides

3 u - u = u - 2 - u

Simplify

2 u = - 2

2 u = - 2

Divide both sides by

2

2 u = - 2

Divide both sides by

2

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

Simplify

u = - \frac{2}{2}

u = - \frac{2}{2}

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

sin (x) = - \frac{2}{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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

tan^{2} (x) = 2 tan (x) sin (x)

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

sin (u) = - \frac{3}{5}

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

What is the general solution for 3sin(x)+sqrt(2)=sin(x) ?
The general solution for 3sin(x)+sqrt(2)=sin(x) is x=(5pi)/4+2pin,x=(7pi)/4+2pin