2009 AMC 12B Problem 24

Question

For how many values of

x

in

[0, \pi]

is

\sin^{-1}(\sin 6x) = \cos^{-1}(\cos x)?

Note: The functions

\sin^{-1} = \arcsin

and

\cos^{-1} = \arccos

denote inverse trigonometric functions.

3

4

5

6

7

On

[0, \pi],\ \cos^{-1}(\cos x) = x,

and the left side lies in

[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],

so any solution needs

x \in [0, \tfrac{\pi}{2}]

Since

\sin

is one-to-one on

[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],

the equation reduces to

\sin 6x = \sin x

on

[0, \tfrac{\pi}{2}]

On

[0, \pi],\ \cos^{-1}(\cos x) = x.

Since

\sin^{-1}

takes values in

[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],

any solution requires

x \in [0, \tfrac{\pi}{2}],

where the equation becomes

\sin 6x = \sin x.

The equation

\sin6x=\sin x

holds exactly when

\begin{aligned} 6x&=x+2k\pi, \\ \text{or}\qquad 6x&=\pi-x+2k\pi \end{aligned}

for some integer

k.

The first family gives

x=\tfrac{2k\pi}{5},

contributing

0

and

\tfrac{2\pi}{5}.

The second gives

x=\tfrac{(2k+1)\pi}{7},

contributing

\tfrac\pi7

and

\tfrac{3\pi}{7}.

Hence there are

4

solutions in

[0,\tfrac\pi2].

Thus, the correct answer is B.

Problem 24 in other years