Written by Colin+ in ask uncle colin, trigonometry.

Dear Uncle Colin,

I'm trying to solve $2\cos(3x)-3\sin(3x)=-1$ (for $0\le \theta \lt 90º$) 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 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 $0 \le \theta \lt 90º$.

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

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 $65.9º$.

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, $\sin(3x) = \frac{6\pm\sqrt{36+4(13)(3)}}{26} = \frac{ 6\pm\sqrt{192}}{26}$.

*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*

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

Since $0 \le x \lt 90º$, $0 \le 3x \lt 270º$.

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