Question Number 181521 by CrispyXYZ last updated on 26/Nov/22
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\begin{cases}{\mathrm{sin}{x}>\mathrm{0}}\\{\sqrt{\mathrm{3}}\mathrm{sin}{x}+\mathrm{cos}{x}>\mathrm{0}}\\{\mathrm{0}<{x}<\mathrm{2}\pi}\end{cases} \\ $$
Answered by mr W last updated on 26/Nov/22
$$\mathrm{sin}\:{x}\:>\mathrm{0}\:\Rightarrow\mathrm{0}<{x}<\pi \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}>\mathrm{0}\:\Rightarrow\mathrm{cot}\:{x}>−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{0}<{x}<\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$
Answered by mahdipoor last updated on 26/Nov/22
$$\sqrt{\mathrm{3}}{sinx}+{cosx}=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}+\frac{\mathrm{1}}{\mathrm{2}}{cosx}\right)= \\ $$$$\mathrm{2}\left({cos}\mathrm{30}{sinx}+{sin}\mathrm{30}{cosx}\right)=\mathrm{2}{sin}\left(\mathrm{30}+{x}\right)>\mathrm{0} \\ $$$$\Rightarrow\mathrm{360}{n}<\mathrm{30}+{x}<\mathrm{360}{n}+\mathrm{180} \\ $$$$\Rightarrow{sinx}>\mathrm{0}\Rightarrow\mathrm{360}{n}<{x}<\mathrm{360}{n}+\mathrm{180} \\ $$$$\Rightarrow\mathrm{0}<{x}<\mathrm{360} \\ $$$$\Rightarrow\Rightarrow\Rightarrow\mathrm{0}<{x}<\mathrm{150}\:{or}\:\mathrm{0}<{x}<\frac{\mathrm{5}}{\mathrm{6}}\pi \\ $$