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

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

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cos(-x)=-cos(x)

cos (- x) = - cos (x)

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

Degrees

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

Solution steps

cos (- x) = - cos (x)

Rewrite using trig identities

cos (- x) = - cos (x)

cos (x) = - cos (x)

cos (x) = - cos (x)

Subtract

- cos (x)

from both sides

2 cos (x) = 0

Divide both sides by

2

2 cos (x) = 0

Divide both sides by

2

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

Simplify

cos (x) = 0

cos (x) = 0

General solutions for

cos (x) = 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}

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

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

9 sin (x) tan (x) = - 4 tan (x)

tan^{2} (θ) - 3 tan (θ) = 0

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

2 cos^{2} (x) + 3 cos (x) = 0

y, sin (y) = x

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