# Solving a set of 3 Nonlinear Equations In the following 3 equations: k 1 </msub>

Solving a set of 3 Nonlinear Equations
In the following 3 equations:
${k}_{1}{\mathrm{cos}}^{2}\left(\theta \right)+{k}_{2}{\mathrm{sin}}^{2}\left(\theta \right)={c}_{1}$
$2\left({k}_{2}-{k}_{1}\right)\mathrm{cos}\left(\theta \right)\mathrm{sin}\left(\theta \right)={c}_{2}$
${k}_{1}{\mathrm{sin}}^{2}\left(\theta \right)+{k}_{2}{\mathrm{cos}}^{2}\left(\theta \right)={c}_{3}$
${c}_{1}$, ${c}_{2}$ and ${c}_{3}$ are given, and ${k}_{1}$, ${k}_{2}$ and $\theta$ are the unknowns. What is the best way to solve for the unknowns?
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isscacabby17
Re-write in terms of double-angle expressions, and define values $A$, , $C$:
$\begin{array}{rlr}{k}_{1}{\mathrm{cos}}^{2}\theta +{k}_{2}{\mathrm{sin}}^{2}\theta ={c}_{1}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{2{c}_{1}-{k}_{1}-{k}_{2}}{{k}_{1}-{k}_{2}}& =:A\\ 2\left({k}_{1}-{k}_{2}\right)\mathrm{sin}\theta \mathrm{cos}\theta ={c}_{2}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{sin}2\theta =\frac{{c}_{2}}{{k}_{1}-{k}_{2}}& =:B\\ {k}_{1}{\mathrm{sin}}^{2}\theta +{k}_{2}{\mathrm{cos}}^{2}\theta ={c}_{3}& \phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{cos}2\theta =\frac{{k}_{1}+{k}_{2}-2{c}_{3}}{{k}_{1}-{k}_{2}}& =:C\end{array}$
Now, $A=C$ implies ${k}_{1}+{k}_{2}={c}_{1}+{c}_{3}$ (but we'd know that simply by adding your first and third equations together), so that ${k}_{2}={c}_{1}+{c}_{3}-{k}_{1}$ and
$A=\frac{{c}_{1}-{c}_{3}}{2{k}_{1}-{c}_{1}-{c}_{3}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B=\frac{{c}_{2}}{2{k}_{1}-{c}_{1}-{c}_{3}}$
Observe that you can already write
$\mathrm{tan}2\theta =\frac{{c}_{2}}{{c}_{1}-{c}_{3}}$
which gives you $\theta$. For ${k}_{1}$ and ${k}_{2}$, note that ${A}^{2}+{B}^{2}=1$ implies
$\left({c}_{1}-{c}_{3}{\right)}^{2}+{c}_{2}^{2}={\left(2{k}_{1}-{c}_{1}-{c}_{3}\right)}^{2}$
so that
$\begin{array}{rl}{k}_{1}& =\frac{1}{2}\left({c}_{1}+{c}_{3}±\sqrt{\left({c}_{1}-{c}_{3}{\right)}^{2}+{c}_{2}^{2}}\right)\\ {k}_{2}& =\frac{1}{2}\left({c}_{1}+{c}_{3}\mp \sqrt{\left({c}_{1}-{c}_{3}{\right)}^{2}+{c}_{2}^{2}}\right)\end{array}$
###### Not exactly what you’re looking for?
Ciara Mcdaniel
Hint:
${k}_{1}{\mathrm{cos}}^{2}\left(\theta \right)+{k}_{2}{\mathrm{sin}}^{2}\left(\theta \right)={c}_{1}$
${k}_{1}{\mathrm{sin}}^{2}\left(\theta \right)+{k}_{2}{\mathrm{cos}}^{2}\left(\theta \right)={c}_{3}$
${c}_{1}+{c}_{3}={k}_{1}+{k}_{2}$
$2\left({k}_{2}-{k}_{1}\right)\mathrm{cos}\left(\theta \right)\mathrm{sin}\left(\theta \right)={c}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left({k}_{2}-{k}_{1}\right)\mathrm{sin}2\theta ={c}_{2}$
$\left({k}_{2}-{k}_{1}\right)\mathrm{sin}2\theta +{k}_{1}+{k}_{2}={c}_{1}+{c}_{2}+{c}_{3}$
${k}_{2}\left(\mathrm{sin}2\theta +1\right)+{k}_{1}\left(1-\mathrm{sin}2\theta \right)={c}_{1}+{c}_{2}+{c}_{3}$