# Solve: 4 sin(2y-0.3)+5 cos(2y-0.3)=0

Solve: $4\mathrm{sin}\left(2y-0.3\right)+5\mathrm{cos}\left(2y-0.3\right)=0$
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davonliefI
$4\mathrm{sin}\left(x\right)+5\mathrm{cos}\left(x\right)=0$
$4\mathrm{sin}=-5\mathrm{cos}$
$\mathrm{sin}\left(x\right)=\left(-\frac{5}{4}\right)\mathrm{cos}i\left(x\right)$
${\mathrm{sin}}^{2}=\left(\frac{25}{16}\right){\mathrm{cos}}^{2}=1.5625{\mathrm{cos}}^{2}$
${\mathrm{sin}}^{2}=1.5625\cdot \left[1-{\mathrm{sin}}^{2}\left(x\right)\right]$
${\mathrm{sin}}^{2}\left(1+1.5625\right)=1.5625$
$2.5625{\mathrm{sin}}^{2}=1.5625$
${\mathrm{sin}}^{2}\left(x\right)=\left(\frac{1.5625}{2.5625}\right)=0.609756097$
$\mathrm{sin}\left(x\right)=\sqrt{0.609\dots .}=0.780868809$
invers $\mathrm{sin}x=51.34019175$ deg
$2y-0.3=51.34019175$
$2y=51.34+0.3=51.64019175$
$y=\frac{51.640}{2}=25.82009587$