sec^2(x)=4tan(x)

**Furyan** · November 23rd 2012, 03:41 PM

Hello

The question is solve on the interval $-\pi\leq x \leq\pi$

$\sec^2(x) = 4\tan(x)$

using the identity $\sec^2(x) = \tan^2(x) +1$

I get:

$\tan^2(x) - 4\tan(x) +1 = 0$

I can't factor that, so I completed the square and got.

$\tan(x) = 2\pm\sqrt{3}$

It looks like that will give me the solutions, but I've never completed the square on a trigonometric function before and just wanted to check that it's the right way to proceed.

Thank you.

**MarkFL** · November 23rd 2012, 03:46 PM

Yes, what you did is the way to go.

**Furyan** · November 23rd 2012, 03:57 PM

Originally Posted by MarkFL2

Yes, what you did is the way to go.

Thank you MarkFL2,

That's good to know. I'll keep going then

.

**topsquark** · November 23rd 2012, 05:55 PM

If you are solving for x there is another useful identity: the double angle formula.

$sec^2(x) = 4~tan(x)$

$\frac{1}{cos^2(x)} = 4 \cdot \frac{sin(x)}{cos(x)}$

As long as we agree that cos(x) is not zero we may cancel one of the cosine terms:
$\frac{1}{cos(x)} = 4 \cdot sin(x)$

$1 = 4 \cdot sin(x)~cos(x)$

Now, 2 sin(x) cos(x) = sin(2x). Thus

$1 = 2 \cdot sin(2x)$

And go from there. I get that
$x = \{ \frac{\pi}{12},~\frac{5 \pi}{12} \}$

**Furyan** · November 23rd 2012, 06:53 PM

Thank you topsquark,

That's very helpful.

I get the same solutions both ways. I think I need to know the double angle formula and it's really useful to know how to apply in this case

**AZach** · November 23rd 2012, 09:15 PM

Just wondering, how do you complete the square on a trigonometric function? In Furyan's example $\tan^2(x) - 4\tan(x) +1 = 0$ , half of b would be 2, and then squared is 4, but how do you condense $tan^2(x) -4tan(x)+5$ ?

**MarkFL** · November 23rd 2012, 09:25 PM

You could write:

$\tan^2(x)-4\tan(x)=-1$

Add 4 to both sides:

$\tan^2(x)-4\tan(x)+4=3$

$(\tan(x)-2)^2=3$

$\tan(x)-2=\pm\sqrt{3}$

$\tan(x)=2\pm\sqrt{3}$

**topsquark** · November 24th 2012, 06:39 AM

Originally Posted by AZach

Just wondering, how do you complete the square on a trigonometric function? In Furyan's example $\tan^2(x) - 4\tan(x) +1 = 0$ , half of b would be 2, and then squared is 4, but how do you condense $tan^2(x) -4tan(x)+5$ ?

Perhaps it would be easier for you to define y = tan(x). Then we have
$y^2 - 4y + 1 = 0$

And of course once you get the solution for y then you plug back the tan(x) = y.

-Dan

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