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

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

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cos(t)-sin(t)=0

cos (t) - sin (t) = 0

t = \frac{π}{4} + πn

Degrees

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

Solution steps

cos (t) - sin (t) = 0

Rewrite using trig identities

cos (t) - sin (t) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (t) = 0

1 - tan (t) = 0

Move

1

to the right side

1 - tan (t) = 0

Subtract

1

from both sides

1 - tan (t) - 1 = 0 - 1

Simplify

- tan (t) = - 1

- tan (t) = - 1

Divide both sides by

- 1

- tan (t) = - 1

Divide both sides by

- 1

\frac{- tan ( t )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (t) = 1

tan (t) = 1

General solutions for

tan (t) = 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}

t = \frac{π}{4} + πn

t = \frac{π}{4} + πn

sin (x) - cos (x) = \frac{1}{3}

tan^{2} (x) = sec (x) - 1, 0 \leq x \leq 2 π

sin^{2} (θ) = 2 sin^{2} (\frac{θ}{2})

cos (x) = cos (x)

(sec^{2} (x) + 1) (sec^{2} (x) - 1) = tan (x)

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