Solving Trig Equations: Tips, Tricks, and Strategies for All Values of x (0-360)

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To solve the equation 2cot²x + 7cosec x - 13 = 0 for x between 0 and 360 degrees, it's essential to utilize trigonometric identities. A suggested approach is to multiply by sin²x and apply the identity cot²x + 1 = csc²x to simplify the equation. This method can help in transforming the equation into a more manageable form. After applying these strategies, the original poster successfully solved the equation. Understanding and using trigonometric identities is crucial for solving such equations effectively.
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The question is to solve the equation;
2\cot^{2}x + 7cosec x - 13 = 0
for all values of x between 0 and 360. I know I need to use a trig ident, but I've been using several over the course of an hour and I can't seem to get anywhere. Any hints would be helpful
 
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Multiply with sin^2x
 
try using cot(x)^2 + 1 = csc(x)^2

~Lyuokdea
 
Thank's got it.
 
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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