# Cosine Trigonometry Question

Find the radian measure of $\theta$ if $0 \leq \theta \leq 2\pi$ and $$\cos(\theta)(2\cos(\theta)-1) = 0.$$ I'm very new to this topic, so what I did was to take the inverse of $\cos$ from both sides, and then you're left with $$2\cos(\theta) - 1 = 1.570\,\,\, (\text{radians}).$$ Then get rid of the 2 and -1: $\cos(\theta) = 0.285$. Now take the inverse from both sides again and you end up with 1.281 radians.

My calculations are probably complete B.S. Can anyone help me out here?

1. Why are there more than 1 answers on my answer sheet?
2. Can you please help me get the right answers?
3. This is on a non calculator paper - (however my teacher admits that several questions belong in the calculator paper) - Is it likely that the question will be in the calculator paper instead, or am I just missing the big point of it all?

Thanks so much, John.

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So you find the values of $\theta$ within $[0,2\pi]$ that satisfy $\cos\theta=0$ or $\cos\theta=\frac12$... –  Ｊ. Ｍ. Aug 21 '11 at 22:29
...you may want to recall the unit circle and those special angles at that point. –  Ｊ. Ｍ. Aug 21 '11 at 22:32
I've added LaTeX formatting to your question, John. –  Zev Chonoles Aug 21 '11 at 22:40
Possibly the idea of taking the inverse $\cos$ came from the parentheses. The analysis may be more obvious if the equation is written as $(\cos\theta)(2\cos\theta -1)=0$. Or, if we really want parentheses around $\theta$, $[\cos(\theta)][2\cos(\theta)-1]=0$. –  André Nicolas Aug 22 '11 at 6:14

## 1 Answer

Note that, for any two numbers $x$ and $y$, $$xy =0 \qquad\text{ if and only if }\qquad x=0,\text{ or }y=0.$$ Thus, $$\cos(\theta)(2\cos(\theta)-1)=0\qquad\text{ if and only if }\qquad \cos(\theta)=0,\text{ or }(2\cos(\theta)-1)=0.$$ So, there are two possibilities; either $\cos(\theta)=0$, or $2\cos(\theta)-1=0$. Rearranging the second equation, we get that either $$\cos(\theta)=0\qquad\text{ or }\qquad \cos(\theta)=\frac{1}{2}.$$ Do you know which values of $\theta$ are between $0$ and $2\pi$ (i.e. $0\leq \theta\leq 2\pi$) and make $\cos(\theta)=0$? (Look at the graph of $\cos$ if you're not sure.)

What about the values of $\theta$ between $0$ and $2\pi$ that make $\cos(\theta)=\frac{1}{2}$? Note that if $\cos(\theta)=\frac{1}{2}$, then we can make a right triangle with $\theta$ as one of the angles, and
Can you find the value of OPP using the Pythagorean theorem? Do you know what kind of triangle has sides in these proportions? Take a look here.

Your approach of dividing by $\cos(\theta)$ on both sides of the equation $$\cos(\theta)(2\cos(\theta)-1)=0$$ is flawed, because if $\cos(\theta)=0$ then we have just divided by $0$, which isn't allowed. To make this clearer, suppose we are looking for the values of $x$ such that $$x^2-x=0.$$ There are two solutions: $x=0$ and $x=1$. We can't divide both sides by $x$ to get $$x-1=0$$ which has only one solution (namely, $x=1$) because then we will miss the solution $x=0$.

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@John: "Look at the graph of $\cos$ if you're not sure." is one way; the other (if you happen to be more familiar with it) would be to recall that the cosine corresponds to the x-coordinate of a point on the unit circle. Now, you will have to recall some of the special angles your teacher may have asked you to commit to memory... –  Ｊ. Ｍ. Aug 21 '11 at 22:42
Thanks so much! I assume that this is a calculator paper question then, cause I doubt she wants us to remember specific values. On the calculator I can call up the cos graph by typing y=cos(x) right? How do I check for certain values on that graph? Its one of those graphic Texas INstruments Calculators that schools supply. –  John Aug 21 '11 at 22:45
Sorry I have been teaching the whole of precalc to myself in the past 2 weeks. I am doing great on everything else now, but trig is hard without a teacher. –  John Aug 21 '11 at 22:49
If you're using the TI-83 Plus (the one that I recall is pretty common) @John, press the [TRACE] button and then you can enter values for checking... –  Ｊ. Ｍ. Aug 21 '11 at 22:51
@John: You actually should have the values of $\theta$ for which $\cos(\theta)=0$ and $\cos(\theta)=\frac{1}{2}$ memorized; they are pretty important (even better, make sure you understand why it is those values of $\theta$; then if you forget, you can always work it out for yourself). And, personally, I would not treat this is as a calculator question; you can express these $\theta$'s as precise fractions of $\pi$ (like $\frac{\pi}{2}$ or $\frac{2\pi}{3}$, for example), and (in my opinion) you should give your answers like this, not as decimals. –  Zev Chonoles Aug 21 '11 at 22:58