Solve Trig. Problem without a calculator

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The discussion revolves around solving the trigonometric equation sec²(x) = (1 + √3) - (1 - √3)tan(x) without a calculator. Participants derive a quadratic equation in terms of tan(x) by substituting sec²(x) with tan²(x) + 1, leading to tan²(x) + (2 - √3)tan(x) - √3 = 0. The quadratic can be solved using factoring or the quadratic formula. Additionally, the Pythagorean identity is referenced to isolate cos² and facilitate solving the equation. The conversation emphasizes algebraic manipulation and identities to arrive at a solution.
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1. sec^2(x) = (1 + sqrt3) - (1 - sqrt3)*tan (x)



2.sec^2(x) = tan^2(x) +1



3. tan^2(x) + tan(x) + (1 - sqrt3)*tan(x) - sqrt3 = 0
tan^(x)[tan(x) +1 +1 - sqrt3) -sqrt(3) = 0
tan^2(x) + (2 - sqrt3) - sqrt(3) = 0
 
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morr485 said:
1. sec^2(x) = (1 + sqrt3) - (1 - sqrt3)*tan (x)



2.sec^2(x) = tan^2(x) +1



3. tan^2(x) + tan(x) + (1 - sqrt3)*tan(x) - sqrt3 = 0
tan^(x)[tan(x) +1 +1 - sqrt3) -sqrt(3) = 0
tan^2(x) + (2 - sqrt3) - sqrt(3) = 0

sec2(x) - 1 + (1 - sqrt(3))tan(x) - sqrt(3) = 0

Replace sec2(x) - 1 with tan2(x) and you will have a quadratic equation in tan(x), which you can solve by factoring (maybe) or by use of the quadratic formula.
 
For #2, sec^2 will equal 1/cos^2, and tan^2 will equal sin^2/cos^2. If you have learned the Pythagorean Identity (sin^2+cos^2=1) you can isolate cos^2 and solve the left side, which will leave you with 1/1-sin^2. Solving the right side, tan^2 will equal sin^2/cos^2, and you can make 1 equal to sin^2/cos^2, which would equal 2sin^2/cos^2. You can solve this equation to prove the left side.
 
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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