Question Number 141110 by Fikret last updated on 15/May/21
![sin^2 x+sinx=(4/3) , x∈[0,π] equation sum of roots in range?](Q141110.png)
$${sin}^{\mathrm{2}} {x}+{sinx}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:,\:\:\:{x}\in\left[\mathrm{0},\pi\right] \\ $$$${equation}\:\:{sum}\:\:{of}\:{roots}\:{in}\:{range}? \\ $$
Answered by MJS_new last updated on 15/May/21
$${s}^{\mathrm{2}} +{s}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{0}\wedge−\mathrm{1}\leqslant{s}\leqslant\mathrm{1}\:\Rightarrow\:{s}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}} \\ $$$$\mathrm{sin}\:{x}\:=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}}\wedge\mathrm{0}\leqslant{x}\leqslant\pi\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\mathrm{arcsin}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}}\right) \\ $$$${x}_{\mathrm{2}} =\pi−\mathrm{arcsin}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}}\right) \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\pi \\ $$