I am just trying to figure out what formula I would use to solve this equation. The problem is solve $\cos(3\theta)=1/2$; for all $0\leq \theta\leq 360^\circ$. I want to say I would use the double angle formula but I am not positive.


Use the fact that $\cos{x}=\cos{y}\iff x=\pm y+2\pi n$.

So we have $$\cos{3\theta}=\frac{1}{2}=\cos\left(\frac{\pi}{3}\right)\\\implies3\theta=\pm\frac{\pi}{3}+2\pi n\\\implies\theta=\pm\frac{\pi}{9}+\frac{2\pi n}{3}$$


You don't need the double angle formula to solve this one!



$\cos^{-1}(1/2)=60$ (in degrees)



but hang on a second!! there are more solutions within $0\leq \theta\leq 360^\circ$. ... actually this domain must be modified for $3\theta$ since you need the answer to $3\theta$ not just $\theta$ like this:

$3*0\leq 3*\theta\leq 3*360^\circ$

$0\leq 3\theta\leq 1080^\circ$

(this is because the "3" modifies the frequency of the cosine curve. If you need more help with this concept tell me)

You actually have 6 correct answers to this question! one of them is the $20 $ degrees. so...

$3\theta=60$ and so $\theta=20$


$3\theta=60+360$ ... $3\theta=420$ and so $\theta=140$


$3\theta=60+360+360$ ... $3\theta=780$ ...and so $\theta=260$


$3\theta=60+360+360+360+360$ ... $\theta=1140$ which is out of domain

That's 3 solutions. The other three can be brought from how the curve looks. which is $360-60$ which is $300$ degrees


$3\theta=300$ and so $\theta=100$


$3\theta=300+360$ ... $\theta=660$ and so $\theta=220$


$3\theta=300+360+360$ ... $\theta=1020$ and so $\theta=340$


$3\theta=300+360+306+306$ ... $\theta=1380$ out of domain again

So your solutions are "$20, 100, 140, 220, 260$ and $340$"!! I hope I made it through without errors.

Here's a brief proof for the working:



Two ideas might help you here:

  • the algebraic principle of performing inverse operations to isolate the variable.
  • the idea that inverse trig functions (like $cos^{-1}$) are multi-valued - they return a set of possible real numbers rather than a single real number. (i.e., rather than saying $cos^{-1}(1)$ = 0, we say that $cos^{-1}(1)$ = {$2 \pi k, k \in \mathbb{Z}$}.

Apply $cos^{-1}$ to both sides of your equation. This gives two possible equations:

$ \\ 3 \theta = \frac{\pi}{3} + 2 \pi k_1, k_1 \in \mathbb{Z} \\$ or $ \\3 \theta = \frac{2 \pi}{3} + 2 \pi k_2, k_2 \in \mathbb{Z} \\$

Dividing both sides (of both equations) by 3 gives:

$\theta = \frac{\pi}{9} + \frac{2 \pi k_1}{3}, k_1 \in \mathbb{Z}$ or $\theta = \frac{2 \pi}{9} + \frac{2 \pi k_2}{3}, k_2 \in \mathbb{Z} \\$

Notice that the division affects the $+ 2 \pi k$ (periodic) part.

Now find all solutions between 0 and $2 \pi$.


You could apply trigonometric formulae several times to find out that $$\cos 3\theta \equiv 4\cos^3\theta - \cos\theta$$

However, solving $4\cos^3\theta-\cos\theta = \tfrac{1}{2}$ is no simpler, and in fact is much harder, than solving your original equation $\cos3\theta = \tfrac{1}{2}$. Solving this equations is really very easy.

If you want to find all solutions with $0 \le \theta \le 360^{\circ}$ then you need to find all solutions with $0 \le 3\theta \le 1080^{\circ}$. You can do this by plotting a graph, or by drawing a CAST quadrant diagram.

You will find that:

$$3\theta = \ldots,60^{\circ}, 300^{\circ}, 420^{\circ}, 660^{\circ}, \ldots$$

Dividing both sides by $3$ gives

$$\theta = \ldots, 20^{\circ}, 100^{\circ}, 140^{\circ}, 220^{\circ}, \ldots$$


Hint: Let $\phi=3\theta$. What angles $\phi$ between $0^\circ$ and $(3)(360^\circ)$ satisfy $\cos\phi=\frac{1}{2}$?

Remark: What is needed here could be called formulas, though I think of them as facts about the geometry of the cosine function. We switch to radians since you used them in your comment. We use the periodicity of cosine, which can be written as the formula $\cos(x+2k\pi)=\cos x$. The fact that $\cos(\pi/3)=\frac{1}{2}$ is one of those "special angle" facts that at one stage one is expected to remember. We also used the fact that $\cos(2\pi-x)=\cos x$, which is the fact that the cosine function is symmetric about $\pi$.

  • $\begingroup$ I have the 5 solutions which are theta= pie/9, 5pie/9, 7pie/9, 11pie/9, and 13pie/9. I just can't remember what formula I am supposed to use. $\endgroup$ – rick Mar 10 '14 at 20:51
  • $\begingroup$ You seem to have switched to radians. Following my hint we get that $\phi$ is one of $\pi/3$, $5\pi/3$, $7\pi/3$, $11\pi/3$, $13\pi/3$, and $17\pi/3$. Divide by $3$ to get the values of $\theta$. Your list missed $17\pi/9$. $\endgroup$ – André Nicolas Mar 10 '14 at 20:57
  • $\begingroup$ The way I got my list is to give the two solutions $\phi$ in the interval $(0,2\pi)$, then add $2\pi$ to each, then add $2\pi$ to each. $\endgroup$ – André Nicolas Mar 10 '14 at 20:58

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