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

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

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sin(pi+x)=sin(x)

Solution

sin (π + x) = sin (x)

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 (π + x) = sin (x)

Rewrite using trig identities

sin (π + x) = sin (x)

Rewrite using trig identities

sin (π + x)

Use the Angle Sum identity:

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

= sin (π) cos (x) + cos (π) sin (x)

Simplify

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

sin (π) cos (x) + cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

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 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - sin (x)

0 - sin (x) = - 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

Popular Examples

cos(x)=-2/pi

cos (x) = - \frac{2}{π}

2sqrt(3)cos(z)+2sin(z)=0

23 cos (z) + 2 sin (z) = 0

(cos(x)+cos(3x))/(cos^2(x)-sin^2(x))=0

\frac{cos ( x ) + cos ( 3 x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0

sin(x)=2cos^2(x)-1

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

tan(t)=(-5)/(12)

tan (t) = \frac{- 5}{12}

Frequently Asked Questions (FAQ)

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