What is the general solution for 6arccos(x)-3pi=3arccos(x) ?

The general solution for 6arccos(x)-3pi=3arccos(x) is x=-1

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6arccos(x)-3pi=3arccos(x)

6 arccos (x) - 3 π = 3 arccos (x)

x = - 1

Solution steps

6 arccos (x) - 3 π = 3 arccos (x)

Solve by substitution

6 arccos (x) - 3 π = 3 arccos (x)

Let:

arccos (x) = u

6 u - 3 π = 3 u

6 u - 3 π = 3 u : u = π

6 u - 3 π = 3 u

Move

3 π

to the right side

6 u - 3 π = 3 u

Add

3 π

to both sides

6 u - 3 π + 3 π = 3 u + 3 π

Simplify

6 u = 3 u + 3 π

6 u = 3 u + 3 π

Move

3 u

to the left side

6 u = 3 u + 3 π

Subtract

3 u

from both sides

6 u - 3 u = 3 u + 3 π - 3 u

Simplify

3 u = 3 π

3 u = 3 π

Divide both sides by

3

3 u = 3 π

Divide both sides by

3

\frac{3 u}{3} = \frac{3 π}{3}

Simplify

u = π

u = π

Substitute back

u = arccos (x)

arccos (x) = π

arccos (x) = π

arccos (x) = π : x = - 1

arccos (x) = π

Apply trig inverse properties

arccos (x) = π

arccos (x) = a \Rightarrow x = cos (a)

x = cos (π)

cos (π) = - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

= - 1

x = - 1

x = - 1

Combine all the solutions

x = - 1

3 sin (2 x) = sin (x)

cos (b) = \frac{12}{37}

27 (sin (2 t) cos (t) - (sin (t))^{3}) = 0

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

cos (t) = \frac{1}{3}

What is the general solution for 6arccos(x)-3pi=3arccos(x) ?
The general solution for 6arccos(x)-3pi=3arccos(x) is x=-1