Understanding and Solving Trigonometry Equations: Tips and Tricks

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To solve the trigonometric equation sec²X + tan²X = 6, it's important to recognize the relationship between secant and tangent through the Pythagorean identity: sec²X = tan²X + 1. This means that you can substitute tan²X + 1 for sec²X in the equation. After substitution, the equation simplifies to tan²X + 1 + tan²X = 6, leading to 2tan²X + 1 = 6. Solving this will help find the values of X that satisfy the original equation.
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Homework Statement



Hi,
This isn't a homework question, I am having a bit of trouble with some Trig equations. If you are worried that you are answering a homework question please give me the solution to a similar question.

Homework Equations



When we have a trig equation like sec2X + tan2X = 6.

I know that sec is equal to 1/cos, and tan is equal to sin/cos, but what do I do with the square? Do I square all the terms or just the denominator?

The Attempt at a Solution



I have tried many times, to many to list here!

Thanks for your help
Dominic
 
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Hint: is there a Pythagorean identity that relates secant and tangent?
 
Hi Dominic! :wink:
Dragonetti said:
When we have a trig equation like sec2X + tan2X = 6.

I know that sec is equal to 1/cos, and tan is equal to sin/cos, but what do I do with the square? Do I square all the terms or just the denominator?

The whole thing …

tan2x = sin2x/cos2x :smile:

(and you can also use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=18"

sec2x = tan2x + 1, same as 1/cos2x = sin2x/cos2x + cos2x/cos2x :wink:)​
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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