# solve the following equation for all values of x: sin^2x+sinxcosx

solve the following equation for all values of x: ${\mathrm{sin}}^{2}x+\mathrm{sin}x\mathrm{cos}x$
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SoosteethicU
$\mathrm{sin}\left(x\right)\left(\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)\right)$. If this expression equals zero, $\mathrm{sin}\left(x\right)=0$ or $\mathrm{sin}\left(x\right)=-\mathrm{cos}\left(x\right)$, so $\mathrm{tan}\left(x\right)=-1.$
Solutions: $\mathrm{sin}\left(x\right)=0:x=n\left(\pi \right),\mathrm{tan}\left(x\right)=-1:x=\left(4n-1\right)\frac{\pi }{4}$ where n is an integer. x is in radians. To convert to degrees put $\left(\pi \right)=180$: 0, 180, 360, ..., 135, 315, ... for example.
$\mathrm{cos}\left(2x\right)=1-2{\mathrm{sin}}^{2}\left(x\right)$, so ${\mathrm{sin}}^{2}\left(x\right)=\frac{1-\mathrm{cos}\left(2x\right)}{2}$
$\mathrm{sin}\left(2x\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$, so $\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=\left(\frac{1}{2}\right)\mathrm{sin}\left(2x\right)$
${\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=\left(\frac{1}{2}\right)\left(1-\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(2x\right)\right)$. Above solutions apply.