$4\cos x^2 - 4\cos x = 2$, find all solutions in the interval $0^º\leq x\leq 360^º$

Question

$4\cos x^2 - 4\cos x = 2$, find all solutions in the interval $0^º\leq x\leq 360^º$

I'm not sure what I'm overlooking or not doing right but I can't seem to figure it out. I've tried factoring out the $4\cos x$, tried factoring it from standard form...what am I missing?

Thank you for your time and I apologize for being dumb

Answer 1

You have a quadratic so you have to solve the quadratic:

$$ 4cos^2(x) - 4cos(x) = 2 \text{, substitute } X = cos(x) \\ 4X^2 - 4X - 2 = 0 $$

Try to factor, multiply "first" and "last", then find factors which subtract to give the middle term:

$$ 4\cdot-2 = -8 \\ 8 = 2\cdot4, 4 - 2 = 2 \text{, nope} \\ 8 = 1\cdot8, 8 - 1 = 7 \text{, nope} $$

Those are all of the factors, so we cannot factor this. Use the quadratic equation:

$$ X = \frac{4 \pm \sqrt{4^2 - 4\cdot4\cdot-2}}{2\cdot4} \\ X = \frac{4\pm \sqrt{16 + 32}}{8} = \frac{4 \pm\sqrt{48}}{8} \\ X = \frac{4 \pm 4\sqrt{3}}{8} = \frac{1 \pm \sqrt{3}}{2} $$

You can't solve this analytically, all you can do is say:

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

We can discard $\frac{1 + \sqrt{3}}{2}$ because that is greater than 1 (since $\sqrt{3} > 1$, $1 + \sqrt{3} > 2$). This single solution is a negative value which means the two solutions are in the $2^\text{nd}$ and $3^\text{rd}$ quadrants. You can take the inverse cosine (or arc-cosine) to find a numerical value, then you can use the unit circle to find the other solution:

$$ x = acos\left(\frac{1 - \sqrt{3}}{2}\right) \approx 111.47^\circ $$

Indeed this is in the $2^\text{nd}$ quadrant. Now to find the value in the 3rd quadrant, you can subtract this value from $360^\circ$. I would actually visualize it by subtracting from $180^\circ$ (this gives the "principle" angle), then add that angle to $180^\circ$, if you do that you get $180^\circ - \theta + 180^\circ = 360^\circ - \theta$. So the other angle is:

$$ \theta_2 \approx 360^\circ - 111.47^\circ = 248.53^\circ $$

The only two solutions in your range therefore are:

$$ x \approx 111.47^\circ \text{ or } x \approx 248.53^\circ $$

Answer 2

I think you mean $\cos^2 x$. It's quadratic formula for $\cos x$. Solve for it. You get some solutions. Be careful, you have to also invert $\cos x$ to finally get $x$ and at that point, you get more than two solutions per period (except for isolated special cases).

Answer 3

I don't think it's a dumb question at all.

This equation is quadratic in $\cos x$, so you can solve it by factoring or by applying the quadratic formula. First, move everything to one side, and note that we can divide all terms by $2$:

$$4\cos^2 x - 4\cos x - 2=0$$ $$2\cos^2 x - 2\cos x - 1=0$$

We can give $\cos x$ a different name - call it $z$ - and then this is the same as:

$$2z^2 - 2z -1 = 0$$

...which you'll note is not factorable. Applying the quadratic formula, we obtain:

$$z=\frac{2\pm\sqrt{4+8}}{4} = \frac12\pm\frac{\sqrt3}2$$

Since $z=\cos x$, we can say that there are two possibilities. Either $\cos x=\frac12+\frac{\sqrt3}2$, or else $\cos x=\frac12-\frac{\sqrt3}2$. The first option is not possible, because $\frac12+\frac{\sqrt3}2>1$. In the second case, though you can take the inverse cosine to get a value for $x$ that's a second quadrant angle. Then you'll have to use the unit circle to see which other angle has the same cosine.

Does that help?

Answer 4

If you make $y = \cos x$, the equation becomes $ 4 y^2 - 4y - 2 = 0$ which implies $ y = \frac{1 \pm \sqrt3}{2}$. Since $1 + \sqrt 3 > 2$ the only one you have to worry about is $y = \frac{1 - \sqrt 3}{2}$, that is, $\cos x = \frac{1 - \sqrt 3}{2}$. This is not solvable "by hand", so I believe the author of the question made some mistake or you're supposed to use a calculator to solve it. If you meant $y = \cos (x^2)$, then it's pretty much impossible algebraically.