# Trigonometry - Find from $2\cos{x} - \cos{}2x = \dfrac{5}{4}$ all values for x in the interval [$0$°, $360$°]

$$2\cos{x} - \cos{}2x = \dfrac{5}{4}$$

Find all values for $$x$$ in the interval [$$0$$°, $$360$$°]

Going from this equation I arrived at (by way of quadratic addition)

$$\cos{x} = \dfrac{1+\sqrt{2}}{2\sqrt{2}}$$

which I translated to

$$x = 31.3997$$° (approximately)

However, there are three other solutions ($$328$$°, $$81$$° and $$278$$°, all with fractions after the decimal point). How do I find these values, and more generally, how do I find all possible values for $$x$$ between $$0$$° and $$360$$°?

$$2\cos x -(-1+2\cos^2 x)=\frac{5}{4}$$

$$2\cos^2 x-2\cos x +\frac{1}{4}=0$$

And I get:

$$\cos{x} = \frac{2 \pm \sqrt{2}}{4}$$

In general, solving an equation for $\cos x$ and getting $\cos x=a$, your final answer would be something like:

$$x=2k\pi \pm a; k \in \mathbb{Z}$$

And you need to replace $k$ to get the appropriate answers in the interval. For instance, for $k=1$:

$$x=2\pi \pm \cos^{-1} \left ( \frac{2 \pm \sqrt{2}}{4} \right )$$

• I think the solution is $\,\displaystyle{\cos x=\frac{2\pm\sqrt{2}}{4}}$ May 28 '12 at 10:32
• @DonAntonio: Right, thanks. May 28 '12 at 10:35
• @DonAntonio: I didn't mean to make you delete your answer, I was suggesting that you can improve it by adding that part. May 28 '12 at 10:50
• Thanx for that but why would I write the same as you? Better for the OP and for all not too add irrelevant stuff if you already covered the main things. May 28 '12 at 10:53
• @DonAntonio: Very true. Thanks for your understanding. May 28 '12 at 11:07

Using the formula $\cos2x=2\cos^2x-1$ we get the equivalent equation $$2\cos x-2\cos^2x+1=\frac54$$ which simplifies to $$\cos^2x-\cos x+\frac18=0$$ This is a quadratic equation in $\cos x$, so we can solve it the usual way to get two solutions:

$$\cos x=\frac{2\pm\sqrt{2}}{4}\qquad(*)$$

Now, for each of these solutions we get the unique solution lying on the interval $[0°,180°]$ using $\arccos$, so the solutions lying on the interval $[0°,180°]$ are exactly:

$$x_{1,2}=\arccos\left(\frac{2\pm\sqrt{2}}{4}\right)$$

Now, if you remember what the graph of $\cos$ looks like, you'll see, that on $[180°,360°]$, the graph is a mirror image of the graph above $[0°,180°]$. Therefore, we will get two more solutions of (*) by subtracting the two solutions we already have from $360°$, i.e.

$$x_3=360°-x_1\\x_4=360°-x_2$$

This is a complete list of solutions.