What is the inverse of f(x)=9cos(2x)+7,0<= x<= pi/2 ?

The inverse of f(x)=9cos(2x)+7,0<= x<= pi/2 is (arccos((x-7)/9))/(2)

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inverse of f(x)=9cos(2x)+7,0<= x<= pi/2

inverse

f (x) = 9 cos (2 x) + 7, 0 \leq x \leq \frac{π}{2}

\frac{arccos ( \frac{x - 7}{9} )}{2}

Solution steps

Solve for

cos (2 x), y = 9 cos (2 x) + 7

cos (2 x) = \frac{y - 7}{9}

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

2 x = arccos (\frac{y - 7}{9})

Solve for

x, 2 x = arccos (\frac{y - 7}{9})

\frac{arccos ( \frac{y - 7}{9} )}{2}

Substitute

y = x

\frac{arccos ( \frac{x - 7}{9} )}{2}

\frac{1}{5 x + 3}

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

s^{6}

f (x) = x^{3} - 2 + 1

f (x) = \frac{6 x - 5}{x + 1}

What is the inverse of f(x)=9cos(2x)+7,0<= x<= pi/2 ?
The inverse of f(x)=9cos(2x)+7,0<= x<= pi/2 is (arccos((x-7)/9))/(2)