Jaqueline Kirby

2022-06-30

The original question was solve for $\theta$ in $65\mathrm{cos}\left(2\theta \right)-56\mathrm{sin}\left(2\theta \right)-55=0$

Layla Love

Remember the formulas
$\mathrm{cos}2\theta =\frac{1-{\mathrm{tan}}^{2}\theta }{1+{\mathrm{tan}}^{2}\theta },\phantom{\rule{2em}{0ex}}\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }$
but first examine the cases $\theta =\pi /2$ and $\theta =-\pi /2$ that would invalidate the substitution.
We have
$65\mathrm{cos}\pi -56\mathrm{sin}\pi -55=-120\ne 0\phantom{\rule{0ex}{0ex}}65\mathrm{cos}\left(-\pi \right)-56\mathrm{sin}\left(-\pi \right)-55=-120\ne 0$
so the substitution is good and doesn't discard solutions.
Set $t=\mathrm{tan}\theta$ for simplicity, so you get
$65\frac{1-{t}^{2}}{1+{t}^{2}}-56\frac{2t}{1+{t}^{2}}-55=0$
that becomes
$60{t}^{2}+56t-5=0$
$\frac{-14+\sqrt{271}}{30}\phantom{\rule{2em}{0ex}}\frac{-14-\sqrt{271}}{30}$
so you get
$\theta =\mathrm{arctan}\frac{-14+\sqrt{271}}{30}+k\pi \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\theta =\mathrm{arctan}\frac{-14-\sqrt{271}}{30}+k\pi$
Note also that your transformation to
$\mathrm{cos}\theta \left(130\mathrm{cos}\theta -122\mathrm{sin}\theta \right)-10=0$
does not reduce the equation to $130\mathrm{cos}\theta -122\mathrm{sin}\theta -10=0$. From
$a\left(b+c\right)-d=0$
you can't deduce
$b+c-d=0$

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