Hi again all,

sec x tan x - cos x cot x = sin x

I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first?

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- April 10th 2010, 06:28 PMDannyMathTrig Equation with varied trig functions
Hi again all,

sec x tan x - cos x cot x = sin x

I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first? - April 10th 2010, 06:47 PMharish21
- April 10th 2010, 06:56 PMDannyMath
I know the trig identities and I went about trying to apply some but I have trouble knowing which to use and how to continue, I will write down what I have in my book that I stopped at:

(sin x/cos^2x) - cos x (1/tanx) = sin x

from there I tried to apply pythagorean theorem for cos^2x but I just dont have enough experience to work it out further, I did keep trying and I will keep trying but for now I am reviewing my whole term in preparation for final. I really don't mean to ask for answers, I DO want to understand these things and I've done 50 exercises just today on it. - April 10th 2010, 07:05 PMharish21
- April 10th 2010, 07:26 PMDannyMath
I seem to have gotten as far as:

(sin^2 x - (1-sin^2 x)^2)/((1-sin^2 x)(sin x)) = sin x

I got that from using the cos^2 x pythagorean identity and then trying to treat them like regular subtraction for fractions (use common denominator).

This is the last section of the precal course and I will admit I have trouble with trig identities - April 10th 2010, 08:18 PMharish21
- April 10th 2010, 09:00 PMDannyMath
Trying to find all solutions in the interval [0, 2pi).

- April 10th 2010, 09:13 PMharish21
- April 12th 2010, 10:49 AMDannyMath
Haha this problem seems rather convoluted to me but maybe it's because I'm new to it! My answers ended up being:

x = pi/6, 5pi/6, 7pi/6, and 11pi/6

Hope I've come to the right solution.

Thank you! - April 12th 2010, 11:10 AMharish21
- April 12th 2010, 11:13 AMDannyMath
Oh right, I forgot to use the square root right? So then the reference angle would be pi/4 rather than pi/6? If so then I've done all my homework and can start on chemistry final hehe.

x = pi/4, 3pi/4, 5pi/4, 7pi/4

Thanks again! - April 12th 2010, 11:22 AMharish21
- April 12th 2010, 11:31 AMDannyMath
Haha the only thing I really like about Chem is the history behind it. Cheers!