Solving a Trig Problem: Find All Solutions in [0, pi)

[Zach_C]

OK, this is simple because it is supposed to be intuitive but I'm just not getting it today.

Here is an example:

Solve 3tan^2x-1=0
3tan^2x = 1
tan^2x = (1/3)
tanx = (+-)(1/sqrt(3))

//All is understood at this point. It's the sentence coming up.

Because tan x has a period of pi, first find all solutions in the interval [0, pi). These are x = pi/6 and x = 5pi/6.

Ok, I understand the algebra part. But I don't get where the intuitively got x = pi/6?? Could someone explain as my book skips over that step. Thank you in advance.

 

[Leong]

there are four answers. each in one of the four quadrants. quadrant I and III give a positive tangent value while quadrant II and IV give a negative tangent value. unless the interval is limited, i think one should give all the answers.

 

[BobG]

It's because you're 'supposed' to memorize your cosines, sines, and tangents for at least key angles.

Knowing just the sines and cosines is usually enough. All your key angles ( \frac{\pi}{6} , \frac{\pi}{4} , \frac{\pi}{3} ) have 2 as a denominator for both the sines and the cosines. So, knowing you should be able to solve the problem, guessing that this must be a key angle is a pretty good guess. Tangent is sine over cosine, or:

\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} is a pretty good guess (the two's canceled out, which is why you didn't see them when you solved for the tangent)

What's the arcsine of \frac{1}{2}?
What's the arcosine of \frac{\sqrt{3}{2}?

If they both come up to the same angle, you've solved the problem. If they don't, your tangent is for one of the more difficult angles and you're going to have to resort to some of the other trig identities to figure the problem out.

Edit: That's strange. Can you only use so many latex images a day? First, the last image cross linked to a different post. Now I get an error message.