Why Is This Trig Equation So Challenging to Solve?

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The discussion centers around the difficulty of solving the trigonometric equation 0.529cosx + 0.916sinx = 1. The user initially struggles with various methods and substitutions without success, expressing frustration over their inability to find the value of x. A helpful formula is introduced, which simplifies the equation to a more manageable form involving sine and cosine. After re-evaluating the application of the formula, the user successfully finds a solution but is curious about the underlying reasoning. The conversation highlights the challenges of trigonometric equations and the effectiveness of specific mathematical techniques.
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I have spent over an hour trying to solve this equation but I cannot seem to do anything with the equation after I get to the point where

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.529cosx + 0.916sinx = 1

I've tried doing a million other substitutions and methods but I just don't know how I should go about finding x.

Any help appreciated, thanks.
 
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A\sin x+B\cos x=\sqrt{A^2+B^2}\cdot\sin(x+\phi)\,, where \displaystyle \phi = \arctan \left(\frac{B}{A}\right) + \begin{cases}0 & \text{if }A \ge 0, \\\pi & \text{if }A < 0,\end{cases}
 
Thank you for the quick reply but I'm not sure what I need to do with that equation.

I tried substituting values and in the end I got x = -38.23 which is incorrect

Edit: For the love of... it worked! Tried it again and apparently I'd done some of mistake, but that formula is magical! I'd be interested in learning why it works but I've already spent about 2 hours on this question, so I will likely come back here to ask in the future.

Thanks so much!
 
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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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