Dear Uncle Colin,

I have to find the values of $x$, between 0 and $\pi$ inclusive, where $2\cos(x) > \sec(x)$. My answer was $0 \le x \lt \piby 4$, but the answer also includes $\piby 2 \lt x \lt \frac{3}{4}\pi$. I don’t understand why!

Stuck Evaluating Confusing And Nasty Trig

Hi, SECANT, and thanks for your message!

What I suspect you’ve done is multiply both sides by $\cos(x)$ to get $2\cos^2(x) > 1$, said $\cos(x) > \frac{1}{\sqrt{2}}$ when $0 \le x \lt \piby 4$.

That’s true, so far as it goes, but it’s not the whole story. Multiplying both sides by $\cos(x)$ is fraught with danger, because it is not guaranteed to be positive.

### What you should do instead

Instead, I would bring everything to the left-hand side:

$2\cos(x) - \sec(x) > 0$

Then turn it into a single fraction:

$\frac{2\cos^2(x) - 1}{\cos(x)} > 0$

Now, look for sign changes: when $\cos^2(x) = \frac{1}{2}$, the fraction is 0 and we have a change of sign. This occurs when $x= \frac{1}{2}\pi$ or $x = \frac{3}{4}\pi$.

We also have a change of sign when $\cos(x)=0$, at $x=\piby 2$. This gives us three interesting points to consider: $\piby 4$, $\piby2$ and $\frac{3}{4}\pi$.

Between $0$ (inclusive) and $\piby 4$ (exclusive), the inequality holds true. It is untrue between $\piby 4$ and $\piby 2$, undefined at $x=\piby2$ and then true until $\frac{3}{4}\pi$ (which is excluded).

The final answer is $0 \le x \lt \piby 4$ or $\piby 2 \lt x \lt \frac{3}{4}\pi$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.