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? 


*

*Why are there more than 1 answers on my answer sheet? 

*Can you please help me get the right answers?

*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.
 A: 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 $x=0$.
