What is the general solution for sin(pi-x)+cos(x)=0 ?

The general solution for sin(pi-x)+cos(x)=0 is x=(3pi)/4+pin

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Popular >

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

Solution

sin (π - x) + cos (x) = 0

Solution

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (π - x) + cos (x) = 0

Rewrite using trig identities

sin (π - x) + cos (x) = 0

Rewrite using trig identities

sin (π - x)

Use the Angle Difference 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))

Refine

= sin (x)

= sin (x)

sin (x) + cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{sin ( x ) + cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

\frac{sin ( x )}{cos ( x )} + 1 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) + 1 = 0

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Popular Examples

sin(x)=-sin(2x)

sin (x) = - sin (2 x)

-3=1+4tan(θ)

- 3 = 1 + 4 tan (θ)

2cos(2θ)-3cos(θ)-2=0

2 cos (2 θ) - 3 cos (θ) - 2 = 0

tan(2θ)+6tan(θ)+8=0

tan (2 θ) + 6 tan (θ) + 8 = 0

cos(φ)=sqrt(3)sin(φ)

cos (φ) = 3 sin (φ)

Frequently Asked Questions (FAQ)

What is the general solution for sin(pi-x)+cos(x)=0 ?
The general solution for sin(pi-x)+cos(x)=0 is x=(3pi)/4+pin