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

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

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acos(x)=cos(x)

a 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

a cos (x) = cos (x)

Solve by substitution

a cos (x) = cos (x)

Let:

cos (x) = u

a u = u

a u = u : u = 0

a u = u

Move

u

to the left side

a u = u

Subtract

u

from both sides

a u - u = u - u

Simplify

a u - u = 0

a u - u = 0

Factor

a u - u : u (a - 1)

a u - u

Factor out common term

u

= u (a - 1)

u (a - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0

Substitute back

u = cos (x)

cos (x) = 0

cos (x) = 0

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

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

Combine all the solutions

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

4674.04 = \frac{5}{3 cos ( x )} \cdot 2804.42

sin (A) + 3 cos (A) = 0

x, 3 cot (x) - 1 = 0

- sin (x) = 3 cos (x)

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

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