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

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

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cos(x)-sin(x)=0,-pi<= x<= pi

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

x = \frac{π}{4}, x = - \frac{3 π}{4}

Degrees

x = 4 5^{\circ}, x = - 13 5^{\circ}

Solution steps

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

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

Solutions for the range

- π \leq x \leq π

x = \frac{π}{4}, x = - \frac{3 π}{4}

5 - tan^{2} (x) = 4

2 sec^{2} (θ) - 3 = tan^{2} (θ)

sin (π - x) + cos (x) = 0

sin (x) = - sin (2 x)

- 3 = 1 + 4 tan (θ)

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