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

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

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-cos(t)=cos(t)

- cos (t) = cos (t)

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

Degrees

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

Solution steps

- cos (t) = cos (t)

Solve by substitution

- cos (t) = cos (t)

Let:

cos (t) = u

- u = u

- u = u : u = 0

- u = u

Move

u

to the left side

- u = u

Subtract

u

from both sides

- u - u = u - u

Simplify

- 2 u = 0

- 2 u = 0

Divide both sides by

- 2

- 2 u = 0

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{0}{- 2}

Simplify

u = 0

u = 0

Substitute back

u = cos (t)

cos (t) = 0

cos (t) = 0

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

General solutions for

cos (t) = 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}

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

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

Combine all the solutions

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

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

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

- 5 π sin (\frac{π ( t )}{2}) = 7.071

9.8 \cdot sin (x) - 19.6 \cdot cos (x) = 4.7

cos (x) = - 3 sin (x)

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