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

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

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

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

x = \frac{π}{4} + πn

Degrees

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

Solution steps

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

Switch sides

- 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

sin (3 θ) = \frac{1}{2}

- sin^{2} (θ) + cos^{2} (θ) = 1 - sin (θ)

1 - sin (θ) = 2 cos^{2} (θ)

2 sin^{2} (θ) - 3 sin (θ) = 0

sec^{2} (x) + 2 tan (x) = 4

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