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Solve the equation for \\(\\theta\\) and \\(x\\)

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\\(\\theta\\) = [[0]]

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\\(x\\) = [[1]]

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\\(\\var{a}e^{j\\theta}+\\var{b}e^{-j\\theta}=\\var{c}+xj\\)

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\\(\\var{a}(cos(\\theta)+jsin(\\theta))+\\var{b}(cos(-\\theta)+jsin(-\\theta))=\\var{c}+xj\\)

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\\(\\var{a}(cos(\\theta)+jsin(\\theta))+\\var{b}(cos(\\theta)-jsin(\\theta))=\\var{c}+xj\\)

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\\(\\var{a}cos(\\theta)+j\\var{a}sin(\\theta))+\\var{b}cos(\\theta)-j\\var{b}sin(\\theta))=\\var{c}+xj\\)

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\\(\\simplify{{a}+{b}}cos(\\theta)+j\\simplify{{a}-{b}}sin(\\theta)=\\var{c}+xj\\)

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This equation can be split into two parts. Real = Real  and Imaginary = Imaginary.

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\\(\\simplify{{a}+{b}}cos(\\theta)=\\var{c}\\)   and   \\(\\simplify{{a}-{b}}sin(\\theta)=x\\)

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\\(cos(\\theta)=\\simplify{{c}/({a}+{b})}\\)   

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\\(\\theta = \\var{theta}\\)

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Inserting this into the second part gives

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\\(\\simplify{{a}-{b}}sin(\\var{theta})=x\\)

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\\(x=\\simplify{({a}-{b})}\\simplify{sin({theta})}\\)

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\\(x=\\var{x}\\)

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Euler's identity \\(e^{j\\theta}=cos(\\theta)+jsin(\\theta)\\)

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Given the equation \\(\\var{a}e^{j\\theta}+\\var{b}e^{-j\\theta}=\\var{c}+xj\\)

", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}