# Ask Uncle Colin: Too Many Answers

Dear Uncle Colin,

I have to solve $5\sin(x) - 5\cos(x)=2$ for $0 \le x \lt 360º$ – I get four answers, but apparently only two are correct. Any suggestions?

Blooming Impeded: Rudimentary Knowledge And Reasoning

Hi, BIRKAR, and thanks for your message!

This looks tasty! Let’s work through it.

Your approach was to set $\cos(x) = \sqrt{1 - \sin^2(x)}$, which is interesting – although a bit dicey. I think it should be $\pm$ the square root, and you need to take inordinate care with the quadrants. I would counsel against doing it that way.

If you wanted to use the $\sin^2(x) + \cos^2(x) \equiv 1$ identity, I would suggest a “rearrange and square” approach.

- $5\sin(x) - 2 = 5\cos(x)$ (You got here by a circuitous route)
- $25\sin^2(x) - 20\sin(x) + 4 = 25 - 25\sin^2(x)$
- $50\sin^2(x) - 20\sin(x) - 21 = 0$

It’s absolutely fine to wallop this with the formula, but I’m going to take a different approach: let $s = 10\sin(x)$ and double everything:

- $100\sin^2(x) - 40\sin(x) - 42 = 0$
- $s^2 - 4s - 42 = 0$

That doesn’t factorise, but the square completes easily: $(s-2)^2 - 46 = 0$

So $s = 2 \pm \sqrt{46}$. We made $s$ up, so going backwards:

$\sin(x) = \frac{2 \pm \sqrt{46}}{10}$

This gives (as you got) four answers for $x$: about 61º and 119º for the positive root and 209º and 331º for the negative one.

However, two of these don’t work: it’s **always** important to check that your answers satisfy the original equation. It turns out that 119º and 331º are spurious solutions.

### Where did they come from?

When you squared the equation, you turned negatives positive – and that’s introduced places where $5\sin(x) - 2 = -5\cos(x)$ as solutions.

### How else could you do it?

I think the most reliable approach here is the harmonic transformation:

$5 \sin(x) - 5 \cos(x) \equiv R \sin(x-\alpha)$, for some $R$ and $\alpha$.

Now, $R\sin(x-\alpha) \equiv R\sin(x)\cos(\alpha) - R\cos(x)\sin(\alpha)$

This means $5 = R\cos(\alpha)$ and $5 = R\sin(\alpha)$.

A bit of trigonometry gives $\alpha = 45º$ and $R = 5\sqrt{2}$.

So: $5\sqrt{2} \sin(x - 45º) = 2$

$\sin(x-45º) = \frac{\sqrt{2}}{5}$

When we un-sine this, we get a principal solution of about 16º. We expect a second solution as well: if $0 \le x \lt 360º$, then $-45º \le x-45º \lt 315º$, so $x-45º$ could also be 164º.

Adding the 45º back gives 61º and 209º, the two correct solutions.

Hope that helps!

- Uncle Colin

- Edited to correct LateX, 2021-12-08.