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

The general solution for cos(θ)-sin(-θ)=0 is θ=(3pi)/4+pin

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cos(θ)-sin(-θ)=0

cos (θ) - sin (- θ) = 0

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) - sin (- θ) = 0

Rewrite using trig identities

cos (θ) - sin (- θ) = 0

cos (θ) - (- sin (θ)) = 0

Simplify

cos (θ) - (- sin (θ)) : cos (θ) + sin (θ)

cos (θ) - (- sin (θ))

Apply rule

- (- a) = a

= cos (θ) + sin (θ)

cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

1 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 + tan (θ) = 0

1 + tan (θ) = 0

Move

1

to the right side

1 + tan (θ) = 0

Subtract

1

from both sides

1 + tan (θ) - 1 = 0 - 1

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

cos^{2} (3 x) = \frac{1}{4}

cos (7 x - 30) sec (x) = 1

sin (θ) = \frac{45}{53}

2 - sin^{2} (θ) + 2 sin (θ) = 0

tan (θ) = \frac{37.3}{36.46}

What is the general solution for cos(θ)-sin(-θ)=0 ?
The general solution for cos(θ)-sin(-θ)=0 is θ=(3pi)/4+pin