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

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

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0=cos(x)-sin(x)

0 = cos (x) - sin (x)

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

0 = cos (x) - sin (x)

Switch sides

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

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

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 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{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = 4

2 sin (x) = sin (2 x)

sin (2 x) cos (x) = 0, 0 \leq x \leq 2 π

tan (x + π) + 2 sin (x + π) = 0

sin (x) = \frac{3}{2}

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