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

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

u + 2 = - u

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

u + 2 = - u

Move

2

to the right side

u + 2 = - u

Subtract

2

from both sides

u + 2 - 2 = - u - 2

Simplify

u = - u - 2

u = - u - 2

Move

u

to the left side

u = - u - 2

Add

u

to both sides

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

cos (x) = \frac{12}{13}

cos (x) (2 sin (x) + 1) = 0

cos (y) = 0

2 sin^{2} (x) - 3 sin (x) = - 1

3 cos^{2} (θ) + 2 sin (θ) - 1 = 0

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