Solve: 3\sin 2x+4 \cos 2x− 2\cos x+6\sin x-6=0 My Try 6 \sin

Jason Yuhas 2021-12-31 Answered
Solve: \(\displaystyle{3}{\sin{{2}}}{x}+{4}{\cos{{2}}}{x}−{2}{\cos{{x}}}+{6}{\sin{{x}}}-{6}={0}\)
My Try
\(\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}\)

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Expert Answer

Timothy Wolff
Answered 2022-01-01 Author has 1852 answers
\(\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}\)
\(\displaystyle{6}{\sin{{x}}}{\cos{{x}}}−{9}{{\sin}^{{2}}{x}}−{{\cos}^{{2}}{x}}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{1}={0}\)
\(\displaystyle-{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}+{2}{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}−{1}={0}\)
\(\displaystyle{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}−{2}{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}+{1}={0}\)
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Philip Williams
Answered 2022-01-02 Author has 3715 answers
x=2y we get after some manipulation
\(\displaystyle-{4}{{\cos}^{{4}}{y}}−{36}{{\cos}^{{2}}{y}}{{\sin}^{{2}}{y}}+{24}{{\cos}^{{3}}{y}}{\sin{{y}}}={0}\)
\(\displaystyle-{4}{{\cos}^{{2}}{y}}{\left({\cos{{y}}}−{3}{\sin{{y}}}\right)}^{{2}}={0}\)
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Vasquez
Answered 2022-01-09 Author has 8850 answers

Abbreviating \(\sin x\) and \(\cos x\) to s and c, and writing \(\sin 2x=2sc\) and \(\cos 2x=2c^2-1\), the equation becomes
\(6sc+4(2c^2-1)-2c+6s-6=0\)
Factoring out a 2 and grouping terms, we rewrite this as
\(3s(c+1)+4c^2-c-5=0\)
But \(4c^2-c-5=(4c-5)(c+1)\), so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to \(\sin x\) and \(\cos x\) and recognizing the Pythagorean triple \(3^2+4^2=5^2\), these become
\(\cos x=-1\) and \(\sin(x+\theta)=1\)
where \(\sin \theta=\frac 45\) and \(\cos \theta=\frac 35\), e.g. \(\theta=\arcsin(\frac 45)\). So the solution set is
\(\{\pi+2n\pi | n\in \mathbb{Z}\} \cup \{ \frac{\pi}{2}−\arcsin(\frac 45)+2n\pi | n \in \mathbb{Z}\}\)

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asked 2021-12-28

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Let the integral be:
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