Strategy for solving $\sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1$

Question

I believe the way to solve the following equation is to use the "R-formula":

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

If so, it should be rewritten as:

$ R\cos(x-\mathcal{L})$ or $R(\cos x\cos\mathcal{L}+\sin x\sin\mathcal{L})$.

With coefficients equated as:

$ \sqrt{3} = R\cos\mathcal{L} $

$ ? = R\sin x $

In the top equation, it looks like $ \cos x $ is the coefficient of $ \sin x $. But that doesn't work with R-formula (does it?).

I have spent some time trying to use double-angle identities for $\cos(2x)$ to remove something from the $\sin(x)\cos(x)$ expression, but am uncertain whether that's possible due to the leading $\sqrt{3}$.

I have also taken several suggestions from threads like this one, but repeatedly end up uncertain due to the same double-angle issue and/or coefficient issue.

Is R-formula the right approach to solving this equation? Should a double-angle identity be used first? Or is an entirely different approach correct?

Answer 1

One more answer won't do any harm, hopefully. Here's another way to look at this "thing". Once you replaced $\sin x \cos x$ with $\frac{1}{2}\sin 2x$, your equation becomes $$\sqrt 3 \cos 2x - \frac{1}{2}\sin 2x = 1.$$ Recalling that cosine and sine are just abscissa and ordinate of points on the circumference of radius $1$, you can find the solutions by intersecting the line $$r: \sqrt 3 X - \frac{1}{2}Y = 1$$ with the circumference $$X^2 + Y^2 = 1.$$ This leads to the system of equations $$ \begin{cases} \sqrt 3 X - \frac{1}{2}Y = 1\\ X^2 + Y^2 = 1. \end{cases} $$

Replacing $Y = 2\sqrt 3 X - 2$ in the second equation yields the quadratic equation $$13X^2-8\sqrt 3 X +3 =0$$ with solutions $$ \cos 2x = X = \frac{4\sqrt 3 \pm 3}{13}.$$ Since solutions in terms of $2x$ are in first and fourth quadrant we can write the first set of solutions as $$2x = \arccos\frac{4\sqrt 3 + 3}{13} + 2k\pi, $$ that is $$\boxed{x = \frac{1}{2}\arccos\frac{4\sqrt 3 + 3}{13} + k\pi, \ \ k \in \Bbb Z}$$ and the second set of solutions as $$2x = -\arccos\frac{4\sqrt 3 - 3}{13} + 2k\pi,$$ that is $$\boxed{x = -\frac{1}{2}\arccos\frac{4\sqrt 3 - 3}{13} + k\pi,\ \ k \in \Bbb Z} $$

Answer 2

Hint: Substitute $$\sin(x)=\frac{2t}{1+t^2},\cos(x)=\frac{1-t^2}{1+t^2}$$ and use that $$\cos(2x)=2\cos^2(x)-1=1-2\sin^2(x)$$

Answer 3

The left-hand side is $\sqrt{3}\cos 2x-\frac{1}{2}\sin 2x=\frac{\sqrt{13}}{2}\cos (2x-\arcsin\frac{1}{\sqrt{13}})$, so you want to solve $\cos (2x-\arcsin\frac{1}{\sqrt{13}})=\frac{2}{\sqrt{13}}$. One solution is $$\arcsin\frac{1}{\sqrt{13}}+\arccos\frac{2}{\sqrt{13}}=\arcsin\frac{1}{\sqrt{13}}+\arcsin\frac{3}{\sqrt{13}};$$I'll leave you to find others, writing them as you wish.

Answer 4

As pointed out in a comment, $\sin x\cos x=\frac{1}{2}\sin(2x)$, so $$1=\sqrt{3}\cos(2x)-\sin x\cos x=(\sqrt3,1/2)\cdot (\cos(2x),\sin(2x)).$$ Now, use the formula for $\mathbb R^2$ scalar product of two vectors to get $$(\sqrt3,1/2)\cdot (\cos(2x),\sin(2x)) = \sqrt{3+1/4}\cos(\alpha(x))=\frac{\sqrt{13}}{2},$$ where $\alpha(x)$ is the angle formed by the vectors $\vec u=(\sqrt3,1/2)$ and $\vec v=(\cos(2x),\sin(2x))$.

Now you have $\cos(\alpha(x))=\frac{2}{\sqrt{13}}$, hence, $\alpha(x)=\arccos(2/\sqrt{13})$.

So, $(\cos(2x),\sin(2x))=(\sqrt{13},1/2)\begin{pmatrix}\cos(\alpha(x))&\sin(\alpha(x))\\-\sin(\alpha(x))&\cos(\alpha(x))\end{pmatrix}$, and with a little bit of calculation, you are able to solve it for $x$.