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

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

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sin(x)=sin(x+(-3)pi)

Solution

sin (x) = sin (x + (- 3) π)

Solution

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) = sin (x + (- 3) π)

Rewrite using trig identities

sin (x) = sin (x + (- 3) π)

Rewrite using trig identities

sin (x + (- 3) π)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (x) cos ((- 3) π) + cos (x) sin ((- 3) π)

Simplify

sin (x) cos ((- 3) π) + cos (x) sin ((- 3) π) : - sin (x)

sin (x) cos ((- 3) π) + cos (x) sin ((- 3) π)

Remove parentheses:

(- a) = - a

= sin (x) cos (- 3 π) + cos (x) sin (- 3 π)

sin (x) cos (- 3 π) = - sin (x)

sin (x) cos (- 3 π)

cos (- 3 π) = - 1

cos (- 3 π)

Use the following property:

cos (- x) = cos (x)

cos (- 3 π) = cos (3 π)

= cos (3 π)

cos (3 π) = cos (π)

cos (3 π)

Rewrite

3 π

2 π + π

= cos (2 π + π)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + π) = cos (π)

= cos (π)

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

= - 1

= (- 1) sin (x)

Refine

= - sin (x)

= - sin (x) + sin (- 3 π) cos (x)

cos (x) sin (- 3 π) = 0

cos (x) sin (- 3 π)

sin (- 3 π) = 0

sin (- 3 π)

Use the following property:

sin (- x) = - sin (x)

sin (- 3 π) = - sin (3 π)

= - sin (3 π)

sin (3 π) = sin (π)

sin (3 π)

Rewrite

3 π

2 π + π

= sin (2 π + π)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + π) = sin (π)

= sin (π)

= - sin (π)

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

= - 0

Simplify

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

= - sin (x) + 0

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

= - sin (x)

= - sin (x)

sin (x) = - sin (x)

sin (x) = - sin (x)

Subtract

- sin (x)

from both sides

2 sin (x) = 0

Divide both sides by

2

2 sin (x) = 0

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{0}{2}

Simplify

sin (x) = 0

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

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Frequently Asked Questions (FAQ)

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