Example 17 from Algebra and Trigonometry, 9.5 Solving Trigonometric Equations

Solving a Multiple Angle Trigonometric Equation

Solve exactly: $\cos (2 x) = \frac{1}{2}$ on $[0, 2 π) .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
We can see that this equation is the standard equation with a multiple of an angle. If $\cos (α) = \frac{1}{2},$ we know $α$ is in quadrants I and IV. While $θ = \cos^{- 1} \frac{1}{2}$ will only yield solutions in quadrants I and II, we recognize that the solutions to the equation $\cos θ = \frac{1}{2}$ will be in quadrants I and IV.
Therefore, the possible angles are $θ = \frac{π}{3}$ and $θ = \frac{5 π}{3} .$ So, $2 x = \frac{π}{3}$ or $2 x = \frac{5 π}{3},$ which means that $x = \frac{π}{6}$ or $x = \frac{5 π}{6} .$ Does this make sense? Yes, because $\cos (2 (\frac{π}{6})) = \cos (\frac{π}{3}) = \frac{1}{2} .$
Are there any other possible answers? Let us return to our first step.

In quadrant I, $2 x = \frac{π}{3},$ so $x = \frac{π}{6}$ as noted. Let us revolve around the circle again:
$\begin{matrix} 2 x & = & \frac{π}{3} + 2 π \\ = & \frac{π}{3} + \frac{6 π}{3} \\ = & \frac{7 π}{3} \end{matrix}$
so $x = \frac{7 π}{6} .$

One more rotation yields

$\begin{matrix} 2 x & = & \frac{π}{3} + 4 π \\ = & \frac{π}{3} + \frac{12 π}{3} \\ = & \frac{13 π}{3} \end{matrix}$
$x = \frac{13 π}{6} > 2 π,$ so this value for $x$ is larger than $2 π,$ so it is not a solution on $[0, 2 π) .$
In quadrant IV, $2 x = \frac{5 π}{3},$ so $x = \frac{5 π}{6}$ as noted. Let us revolve around the circle again:

$\begin{matrix} 2 x & = & \frac{5 π}{3} + 2 π \\ = & \frac{5 π}{3} + \frac{6 π}{3} \\ = & \frac{11 π}{3} \end{matrix}$
so $x = \frac{11 π}{6} .$

One more rotation yields

$\begin{matrix} 2 x & = & \frac{5 π}{3} + 4 π \\ = & \frac{5 π}{3} + \frac{12 π}{3} \\ = & \frac{17 π}{3} \end{matrix}$
$x = \frac{17 π}{6} > 2 π,$ so this value for $x$ is larger than $2 π,$ so it is not a solution on $[0, 2 π) .$
Our solutions are $x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, and \frac{11 π}{6}$ . Note that whenever we solve a problem in the form of $\sin (n x) = c,$ we must go around the unit circle $n$ times.

How did it go?

Solution

Solution