Number of Solutions of $3\cos^2(x)+\cos(x)-2=0$ I'm trying to figure out how many solutions there are for 
$$3\cos^2(x)+\cos(x)-2=0.$$ 
I can come up with at least two solutions I believe are correct, but I'm not sure if there is a third.
 A: Let $y=\cos(x)$. Then the equation becomes
$$3y^2 + y - 2 = 0.$$
This can be solved using the quadratic formula or any other method for solving quadratics. The two solutions are $y=-1$ and $y=\frac{2}{3}$. 
Thus, you want to find solutions to $\cos x = -1$ and $\cos x = \frac{2}{3}$.
Each of those has infinitely many solutions. For $\cos x = -1$, we have that $x$ must be an odd multiple of $\pi$, and every odd multiple of $\pi$ works. That is, $x=(2n+1)\pi$, $n=0,\pm 1,\pm2, \pm 3,\ldots$.
For $\cos x = \frac{2}{3}$, there are two solutions on the interval $[-\pi,\pi]$; they are approximately $x\approx 0.84107$ and $x\approx-0.84107$. Adding $2\pi$ to either one of them will also give solutions.
So you have three infinite families of solutions:
$$\begin{align*}
x &= (2n+1)\pi &n=0,\pm1,\pm 2,\pm 3,\ldots\\
x &\approx 0.84107 + 2n\pi & n=0,\pm1,\pm 2,\pm 3,\ldots\\
x &\approx -0.84107 + 2n\pi & n=0,\pm1, \pm2, \pm3, \ldots
\end{align*}$$
A: We may rewrite the equation as
$$3\cos^2x+\cos x-2=(\cos x+1)(3\cos x-2)=0.$$
The first infinite set of solutions coming from the first factor then is
$$x_1=\pi+2\pi k,k\in\mathbb{Z},$$
while the second factor leads to
$$x_{2,3}=\pm\arccos{\frac{2}{3}}+2\pi k,k\in\mathbb{Z}.$$
A: You have a cuadratic equation it terms of $y = \cos x$. Then 
$$y = \cos x = \frac{{ - 1 \pm 1\sqrt {{1^2} - 4 \cdot 3 \cdot \left( { - 2} \right)} }}{{2 \cdot 3}}=\frac{{ - 1 \pm 5 }}{6}$$
Thus you need
$\cos x = -1$
or
$\cos x = \frac{2}{3}$
The first solution is clearly $(2k+1)\pi$ for any integer $k$. The second solution is not exact and has two values that repeat periodically, namely $x=0.841068..$ and $x=5442116636..$. You can look at the unitary circle to see how they repeat each full turn ($2\pi$).

A: Your equation factors as $(3\cos x-2)(\cos x+1)=0$.
So the solutions to your equation are the solutions to the equations 
$\ \ \ \ \ \  3\cos x-2 =0$ 
and 
$\ \ \ \ \ \  \cos x+1 =0$.
So, the original equation has a solution $x$ if and only if
$$\cos x=  2/3\quad\text{or}\quad \cos x=-1.$$
Restricting our attention to the interval $[0,2\pi]$:
The equation $\cos x=-1$ has the solution  $x=\pi $.
The equation $\cos x= 2/3$ has two solutions, one of which is in the interval $(0, \pi/2)$ and the other in the interval $(3\pi/2, 2\pi)$ (draw the graph of $\cos$ to see this).
This gives a total of three distinct solutions in the interval $[0,2\pi]$.
Of course, as $\cos x$ is $2\pi$-periodic, there are actually infinitely many solutions...



For what it's worth:





Note that if you used the inverse cosine function on your calculator to find a solution to $\cos x=2/3$ by computing $\arccos(2/3)$, it would return only one value; namely, the solution in the interval $[0,\pi]$.
But, given any value of $\alpha\ne-1$, there are two angles in $[0,2\pi]$ whose cosine is $\alpha$. (examine the graph of $\cos$ to see this).  

You should be wary of this sort of thing when solving trigonometric equations.
A: Unfortunately, there are infinitely many solutions. Note, for example, that $\pi$ is a solution. Then we also have that $\pi + 2k \pi$ is a solution for all $k$. But between $0$ and $2\pi$, there are 3 solutions.
