What is the inverse of 8cos(pi^2x-pi^4)-3 ?

The inverse of 8cos(pi^2x-pi^4)-3 is (arccos((x+3)/8)+pi^4)/(pi^2)

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inverse of 8cos(pi^2x-pi^4)-3

inverse

8 cos (π^{2} x - π^{4}) - 3

\frac{arccos ( \frac{x + 3}{8} ) + π ^{4}}{π ^{2}}

Solution steps

Solve for

cos (π^{2} x - π^{4}), y = 8 cos (π^{2} x - π^{4}) - 3

cos (π^{2} x - π^{4}) = \frac{y + 3}{8}

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

π^{2} x - π^{4} = arccos (\frac{y + 3}{8})

Solve for

x, π^{2} x - π^{4} = arccos (\frac{y + 3}{8})

\frac{arccos ( \frac{y + 3}{8} ) + π ^{4}}{π ^{2}}

Substitute

y = x

\frac{arccos ( \frac{x + 3}{8} ) + π ^{4}}{π ^{2}}

What is the inverse of 8cos(pi^2x-pi^4)-3 ?
The inverse of 8cos(pi^2x-pi^4)-3 is (arccos((x+3)/8)+pi^4)/(pi^2)