Ask Uncle Colin: Stuck on some trig

Dear Uncle Colin,

I’m trying to solve $2\cos (3x)-3\sin (3x)=-1$ (for $0\le \theta <90º$) but I keep getting stuck and/or confused! What do you recommend?

- Losing Angles, Getting Ridiculous Answers, Nasty Geometric Equation

Hi, LAGRANGE, and thank you for your message!

There are a couple of ways to approach this: a standard way that I’d recommend, and a slightly different way I think is a bit daft but that I ought to mention.

The standard way

The standard way is to write $2\cos (3x)-3\sin (3x)$ as a single trigonometric function of the form $R\cos (3x+a)$. Because $\cos (3x+a)=\cos (3x)\cos (a)-\sin (3x)\sin (a)$, we can match coefficients and say $R\cos (a)=2$ and $R\sin (a)=3$.

Solving these (for example, by saying $R=\frac{2}{\cos (a)}$ and substituting into the first equation), gives $\tan (a)=\frac{3}{2}$, so $a\approx 56.3º$ and $R=\sqrt{13}$.

You can then solve $\cos (3x+a)=-\frac{1}{\sqrt{13}}$ for $0\le \theta <90º$.

I would recommend changing the domain here: if $0\le \theta <90º$, then $a\le 3\theta +a<270+a$.

The principle value for $3\theta +a$ is 106.1º, which lies in the specified domain. There’s a second solution at (360-106.1)º, or 253.9º, which is also in the domain.

Now to map it back to get $x$: $3x\approx 49.8º$ or $3x\approx 197.6º$, therefore $x\approx 16.6º$ or $65.9º$.

An alternative

I’ve seen it suggested that writing $\cos (3x)=\sqrt{1-{\sin }^2(3x)}$ might be of some use.

That would make the equation $2\sqrt{1-{\sin }^2(3x)}-3sin(3x)=-1$, which (on the plus side) only involves one function, but (on the minus) is a complete mess. Let’s tidy it up:

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

Now square both sides: $4-4{\sin }^2(3x)=9{\sin }^2(3x)-6\sin (3x)+1$

Rearrange: $0=13{\sin }^2(3x)-6\sin (3x)-3$

This is a quadratic in $\sin (x)$. Using the formula, $\sin (3x)=\frac{6\pm \sqrt{36+4(13)(3)}}{26}=\frac{6\pm \sqrt{192}}{26}$.

Is that a…

whoosh

“192 is four less than ${14}^2$, so its square root is 13 less about four-twentyeighths - so about 12 and six sevenths.”

whoosh

Ninja?

So we have, surreptitiously using a calculator, $\sin (3x)\approx 0.7637$ or $\sin (3x)\approx -0.3022$.

Since $0\le x<90º$, $0\le 3x<270º$.

This gives $3x\approx 47.8º$, $3x\approx 130.2º$ or $3x\approx 197.6º$.

Two of those are familiar from the natural way of doing it, but we have an extra ghost solution at $x\approx 43.4º$ - an answer that doesn’t fit the original problem!

The trouble here arrived when we squared everything - this has a habit of introducing spurious answers (it’s a common trick in fake ‘proofs’ that $0=1$ and similar); whenever you square both sides of an equation, it’s best to check you haven’t introduced anything undesirable!

Hope that helps,

- Uncle Colin