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

The general solution for cos(-θ)=-cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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cos(-θ)=-cos(θ)

cos (- θ) = - cos (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (- θ) = - cos (θ)

Rewrite using trig identities

cos (- θ) = - cos (θ)

cos (θ) = - cos (θ)

cos (θ) = - cos (θ)

Subtract

- cos (θ)

from both sides

2 cos (θ) = 0

Divide both sides by

2

2 cos (θ) = 0

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{0}{2}

Simplify

cos (θ) = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin^{2} (x) cos (x) = cos (x)

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

tan (x) + 3 cot (x) = 5 sec (x)

6 sin (θ) + 3 = 0

2 cos (x) + sin (2 x) = 0

What is the general solution for cos(-θ)=-cos(θ) ?
The general solution for cos(-θ)=-cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin