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How to solve the following trigonometric quadratic equation for x

$3 \cos x + r \cos^{2} x - 2 \sin x -r \sin^{2} x = 0$

where r is a constant

Even though this trigonometric quadratic equation has only one variable(i.e. x), it seems that it has two variables due to the two trigonometric ratios (sin x and cos x)

How to proceed in solving the problem?Is it even possible to solve this problem?

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  • $\begingroup$ What can you do with the $\cos^2 x - \sin^2 x$ term? $\endgroup$ – amcalde Oct 21 '15 at 11:11
  • $\begingroup$ it will become $ cos2x $ then what? $\endgroup$ – N.G.Tyson Oct 21 '15 at 11:13
  • $\begingroup$ Wait, sorry. Are you looking for a numerical solution or an analytic one? $\endgroup$ – amcalde Oct 21 '15 at 11:14
  • $\begingroup$ I am looking for an analytic solution. $\endgroup$ – N.G.Tyson Oct 21 '15 at 11:19
  • $\begingroup$ math.stackexchange.com/questions/1490344/homework-problem $\endgroup$ – lab bhattacharjee Oct 21 '15 at 12:50
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Substitute $u = \cos x$. Then you have $$3u+r u^2 - \sqrt{1-u^2} - r(1-u^2) = 0$$ rearrange to get $$(3u+2ru^2-r)^2 = 1 - u^2$$ For a general $r$ you'll have a quartic equation. These can be solved but may not make sense here, given $r$. For example you must have a solution $u$ that is in $[-1,1]$ or it won't make any sense. Also you may need to be careful because we implicitly assumed that $\sin x$ was positive which may be false (in which case your equation changes slightly).

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