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

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

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

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

x = \frac{π}{4} + πn

Degrees

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

Solution steps

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

Rewrite using trig identities

- 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

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

cos (2 θ) = \frac{- 1}{2}

2 sin (3 x) + 3 = 0

cos (4 x) = - \frac{3}{2}

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

tan (2 x) - tan (x) = 0

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