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

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

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

sin (x) 2 = - sin (x)

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\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 = 0

u 2 = - u

Move

u

to the left side

u 2 = - u

Add

u

to both sides

u 2 + u = - u + u

Simplify

u 2 + u = 0

u 2 + u = 0

Factor

u 2 + u : (2 + 1) u

u 2 + u

Factor out common term

u

= u (2 + 1)

(2 + 1) u = 0

Divide both sides by

2 + 1

(2 + 1) u = 0

Divide both sides by

2 + 1

\frac{( 2 + 1 ) u}{2 + 1} = \frac{0}{2 + 1}

Simplify

u = 0

u = 0

Substitute back

u = sin (x)

sin (x) = 0

sin (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

tan^{3} (x) = 1

sin^{2} (x) - 9 cos (x) - 9 = 0

sin (a) = \frac{8}{17}

cot (t) = 1

4 sin (x) + 3 = 5

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