What is the general solution for arccos(2x)=pi ?

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

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arccos(2x)=pi

arccos (2 x) = π

x = - \frac{1}{2}

Solution steps

arccos (2 x) = π

Apply trig inverse properties

arccos (2 x) = π

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

2 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

2 x = - 1

2 x = - 1

Solve

2 x = - 1 : x = - \frac{1}{2}

2 x = - 1

Divide both sides by

2

2 x = - 1

Divide both sides by

2

\frac{2 x}{2} = \frac{- 1}{2}

Simplify

x = - \frac{1}{2}

x = - \frac{1}{2}

x = - \frac{1}{2}

t, x = 4 sin (t)

sin (2 x) + 1.5 cos (x) = 0

sin (w π) = 0

cos (x) = - 8

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

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