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

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

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

inverse

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

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

Solution steps

Solve for

cos (2 x), y = - 7 cos (2 x)

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

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

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

Solve for

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

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

Substitute

y = x

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

lo g_{\frac{5}{4}} (5 x - 4 x^{2})

f (x) = \frac{- 10 x + 1}{9 x + 6}

y = 5 (0.5)^{- x}

f (x) = \frac{( 5 x )}{x + 3}

f (x) = 1 x + 5

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