Rudy Koch

2022-01-23

Solving $\mathrm{sin}2x\mathrm{sin}x+{\mathrm{cos}}^{2}x=\mathrm{sin}5x\mathrm{sin}4x+{\mathrm{cos}}^{2}4x$

oferenteoo

Expert

Use ${\mathrm{cos}}^{2}\left(a\right)+{\mathrm{sin}}^{2}\left(a\right)=1$ first on both sides then factorize.
$\mathrm{sin}\left(x\right)\left[\mathrm{sin}\left(2x\right)-\mathrm{sin}\left(x\right)\right]=\mathrm{sin}\left(4x\right)\left[\mathrm{sin}\left(5x\right)-\mathrm{sin}\left(4x\right)\right]$
Now use $\mathrm{sin}\left(A\right)-\mathrm{sin}\left(B\right)$ as a product on both sides and simplify (giving some roots $\mathrm{sin}\left(\frac{x}{2}\right)=0\right)$
$\mathrm{sin}\left(x\right)\cdot 2\mathrm{cos}\left(\frac{3x}{2}\right)\mathrm{sin}\left(\frac{x}{2}\right)=\mathrm{sin}\left(4x\right)\cdot 2\mathrm{cos}\left(\frac{9x}{2}\right)\mathrm{sin}\left(\frac{x}{2}\right)$
then write $\mathrm{sin}\left(C\right)\mathrm{cos}\left(D\right)$ as a difference of sines on both sides and simplify.
$\frac{12}{\mathrm{sin}\left(\frac{5x}{2}\right)+\mathrm{sin}\left(\frac{-x}{2}\right)}=\frac{12}{\mathrm{sin}\left(\frac{17x}{2}\right)+\mathrm{sin}\left(\frac{-x}{2}\right)}$
Write the difference of sines as a product and solve
$\mathrm{sin}\left(\frac{5x}{2}\right)-\mathrm{sin}\left(\frac{17x}{2}\right)=0$

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