Question 7.7.1: Solving an Equation for a Specified Variable Solve y = 3 cos...

We want to isolate $\cos 2x$ on one side of the equation so that we can solve for 2x, and then for x.

$y = 3 \cos 2x  ←  \fbox{Our goal is to isolate x.}$

$\frac{y}{3} = \cos 2x$               Divide by 3.

$2x = \arccos \frac{y}{3}$           Definition of arccosine

$x = \frac{1}{2} \arccos \frac{y}{3}$           Multiply by $\frac{1}{2}$ .

An equivalent form of this answer is $x = \frac{1}{2} \cos^{-1} \frac{y}{3}$ .

Because the function $y = 3 \cos 2x$ is periodic, with period π, there are infinitely many domain values (x-values) that will result in a given range value ( y-value). For example, the x-values 0 and π both correspond to the y-value 3. See Figure 38. The restriction 0 ≤ x ≤ $\frac{π}{2}$ given in the original problem ensures that this function is one-to-one, and, correspondingly, that

$x =\frac{1}{2} \arccos \frac{y}{3}$

has a one-to-one relationship. Thus, each y-value in [ -3, 3] substituted into this equation will lead to a single x-value.