What is the inverse of f(x)= 3/2 cos(x/3)-1/2 ?

The inverse of f(x)= 3/2 cos(x/3)-1/2 is 3arccos((2x+1)/3)

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inverse of f(x)= 3/2 cos(x/3)-1/2

inverse

f (x) = \frac{3}{2} cos (\frac{x}{3}) - \frac{1}{2}

3 arccos (\frac{2 x + 1}{3})

Solution steps

Solve for

cos (\frac{x}{3}), y = \frac{3}{2} cos (\frac{x}{3}) - \frac{1}{2}

cos (\frac{x}{3}) = \frac{2 y + 1}{3}

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

\frac{x}{3} = arccos (\frac{2 y + 1}{3})

Solve for

x, \frac{x}{3} = arccos (\frac{2 y + 1}{3})

3 arccos (\frac{2 y + 1}{3})

Substitute

y = x

3 arccos (\frac{2 x + 1}{3})

f (x) = a 3 - x

f (x) = f (x) = \frac{1}{x + 5} - 1

y = \frac{1}{x - 2} - 3

f (x) = \frac{( 6 x + 2 )}{x}

sin (0.627)

What is the inverse of f(x)= 3/2 cos(x/3)-1/2 ?
The inverse of f(x)= 3/2 cos(x/3)-1/2 is 3arccos((2x+1)/3)