What is the inverse of 27cos^2(3x) ?

The inverse of 27cos^2(3x) is (arccos((sqrt(3)sqrt(x))/9))/(3),(arccos(-(sqrt(3)sqrt(x))/9))/(3)

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inverse of 27cos^2(3x)

inverse

27 cos^{2} (3 x)

\frac{arccos ( \frac{3 x}{9} )}{3}, \frac{arccos ( - \frac{3 x}{9} )}{3}

Solution steps

Solve for

cos (3 x), y = 27 cos^{2} (3 x)

cos (3 x) = \frac{3 y}{9}, cos (3 x) = - \frac{3 y}{9}

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

3 x = arccos (\frac{3 y}{9}), 3 x = arccos (- \frac{3 y}{9})

Solve for

x, 3 x = arccos (\frac{3 y}{9})

Solve for

x, 3 x = arccos (- \frac{3 y}{9})

\frac{arccos ( \frac{3 y}{9} )}{3}, \frac{arccos ( - \frac{3 y}{9} )}{3}

Substitute

y = x

\frac{arccos ( \frac{3 x}{9} )}{3}, \frac{arccos ( - \frac{3 x}{9} )}{3}

f (x) = 6 - 2 x

f (x) = lo g_{7} (x^{4})

f (x) = 12 - 4 x

f (x) = 2 x^{2} + 8 x - 1

f (x) = \frac{3 x - 2}{5 - 2 x}

What is the inverse of 27cos^2(3x) ?
The inverse of 27cos^2(3x) is (arccos((sqrt(3)sqrt(x))/9))/(3),(arccos(-(sqrt(3)sqrt(x))/9))/(3)