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Question Number 181521 by CrispyXYZ last updated on 26/Nov/22

Find the range of x  such that   { ((sinx>0)),(((√3)sinx+cosx>0)),((0<x<2π)) :}

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x}\:\:\mathrm{such}\:\mathrm{that} \\ $$ $$\begin{cases}{\mathrm{sin}{x}>\mathrm{0}}\\{\sqrt{\mathrm{3}}\mathrm{sin}{x}+\mathrm{cos}{x}>\mathrm{0}}\\{\mathrm{0}<{x}<\mathrm{2}\pi}\end{cases} \\ $$

Answered by mr W last updated on 26/Nov/22

sin x >0 ⇒0<x<π  (√3) sin x+cos x>0 ⇒cot x>−(√3)  ⇒0<x<((5π)/6)

$$\mathrm{sin}\:{x}\:>\mathrm{0}\:\Rightarrow\mathrm{0}<{x}<\pi \\ $$ $$\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}>\mathrm{0}\:\Rightarrow\mathrm{cot}\:{x}>−\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{0}<{x}<\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

Answered by mahdipoor last updated on 26/Nov/22

(√3)sinx+cosx=2(((√3)/2)sinx+(1/2)cosx)=  2(cos30sinx+sin30cosx)=2sin(30+x)>0  ⇒360n<30+x<360n+180  ⇒sinx>0⇒360n<x<360n+180  ⇒0<x<360  ⇒⇒⇒0<x<150 or 0<x<(5/6)π

$$\sqrt{\mathrm{3}}{sinx}+{cosx}=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}+\frac{\mathrm{1}}{\mathrm{2}}{cosx}\right)= \\ $$ $$\mathrm{2}\left({cos}\mathrm{30}{sinx}+{sin}\mathrm{30}{cosx}\right)=\mathrm{2}{sin}\left(\mathrm{30}+{x}\right)>\mathrm{0} \\ $$ $$\Rightarrow\mathrm{360}{n}<\mathrm{30}+{x}<\mathrm{360}{n}+\mathrm{180} \\ $$ $$\Rightarrow{sinx}>\mathrm{0}\Rightarrow\mathrm{360}{n}<{x}<\mathrm{360}{n}+\mathrm{180} \\ $$ $$\Rightarrow\mathrm{0}<{x}<\mathrm{360} \\ $$ $$\Rightarrow\Rightarrow\Rightarrow\mathrm{0}<{x}<\mathrm{150}\:{or}\:\mathrm{0}<{x}<\frac{\mathrm{5}}{\mathrm{6}}\pi \\ $$

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