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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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    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 $\theta$, $[\cos(\theta)][2\cos(\theta)-1]=0$.2011-08-22

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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 enter image description here
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: 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.2011-08-21