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

The general solution for 3arccos(x)=pi is x= 1/2

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3arccos(x)=pi

3 arccos (x) = π

x = \frac{1}{2}

Solution steps

3 arccos (x) = π

Divide both sides by

3

3 arccos (x) = π

Divide both sides by

3

\frac{3 arccos ( x )}{3} = \frac{π}{3}

Simplify

arccos (x) = \frac{π}{3}

arccos (x) = \frac{π}{3}

Apply trig inverse properties

arccos (x) = \frac{π}{3}

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

x = cos (\frac{π}{3})

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

= \frac{1}{2}

x = \frac{1}{2}

x = \frac{1}{2}

3 sin (2 x) - 2 sin (2 x) = 0

2 cos (4 x) + 5 = 6

cos (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

2 cos^{2} (a) = 1 + cos (12 0^{\circ})

cos (x) = \frac{11}{17 \cdot 38}

What is the general solution for 3arccos(x)=pi ?
The general solution for 3arccos(x)=pi is x= 1/2