$2 \cos^2 x − 2 \cos x− 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$

Question

Find the solutions of $$2 \cos^2 x − 2 \cos x− 1 = 0$$ for all $0^\circ ≤ x < 360^\circ$.

For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.

Is this correct? Thanks!

Answer 1

Observe that $2 \cos^2 x - 2 \cos x - 1$ is irreducible and cannot be factored.

Apply the quadratic formula: $$\cos x = \frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$$ But $\frac{1 + \sqrt{3}}{2}$ is not in between $-1$ and $1$, so we can discard it accordingly. Thus, $$\cos x = \frac{-1-\sqrt{3}}{2}$$ and so $$x=\cos^{-1} \bigg(\frac{-1-\sqrt{3}}{2}\bigg)\approx \boxed{111.5,248.5}$$

Answer 2

You have a quadratic equation if you define $y=\cos(x)$; the solutions corresponds to $\cos(x)=\frac{1}{2} \left(1-\sqrt{3}\right)$ and $\cos(x)=\frac{1}{2} \left(1+\sqrt{3}\right)$; the second solution is larger than $1$ and so it can be discarded. So, you look for the solution of $$ \cos(x)=\frac{1}{2} \left(1-\sqrt{3}\right)$$ for $0 \leq x \leq 2\pi$. The two possible solutions are then (in radians) $x=1.94553$ and $x=4.33765$ that is to say (in degrees) $x=111.471$ and $x=248.529$ which match your results.

Answer 3

Let $y = \cos(x)$. Then we have the following:

$$2y^2 - 2y - 1 = 0$$

Using the quadratic formula, we get the roots to be:

$$y = \cos(x) = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$$

And so:

$$x = \arccos(\frac{1}{2} \pm \frac{\sqrt{3}}{2}) $$

Since $\frac{1}{2} + \frac{\sqrt{3}}{2}>1$, we can ignore this solution. Since both of your solutions satisfy $\cos(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}$, you are indeed correct.