Quadratic Cosine

We can do this by treating the equation like a normal quadratic equation, letting $x = cos θ$ , solving for $x$ , and then seeing at what $θ$ that $x = cos θ$ .

In other words:

Let $x = cos θ$

The equation becomes $2x^2 - 7x + 3 = 0$

To this we can apply the quadratic formula to find the value of $x$ .

$x =$ $\frac{-b \pm \sqrt{b^2-4ac}}{2a} =$ $\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(3)}}{2(2)} =$ $\frac{7 \pm \sqrt{49-24}}{4} =$ $\frac{7 \pm \sqrt{25}}{4} =$ $\frac{7 \pm 5}{4} =$

$3$ and $\frac{1}{2}$

Thus $x = 3$ and $x = \frac{1}{2}$

But $x = cos θ$ , so

$cos θ = 3$ and $cos θ = \frac{1}{2}$

Now we must check both of these to see if there is a solution for $θ$ and, if there is a solution, whether it is it is within $0≤θ≤π$

The value of $cos θ$ is always oscillating between $-1$ and $1$ for all $θ$ . Because it never goes beyond $-1$ and $1$ we will never be able to find a $θ$ to make $cos θ = 3$ . Therefore there is no solution for $cos θ = 3$ .

$cos θ = \frac{1}{2}$ on the other hand does have a solution as it is between $-1$ and $1$ . By observing a graph of $cos θ$ we see that $cos θ = \frac{1}{2}$ someplace between $0$ and $π$ . Thus we have a $θ$ that is a solution and within $0≤θ≤π$ .

This gives us a total of one solution.