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Question Number 59113 by naka3546 last updated on 04/May/19

$$Letaisarealnumber.Howmanysolutions$$ $$cantheequationin\theta$$ $$(\sin \theta +\cos \theta )(\sin \theta \cos \theta -1)=a$$ $$havefor0<\theta <\frac{\pi }{2}?$$

Answered by MJS last updated on 05/May/19

$$(\sin \theta +\cos \theta )(\sin \theta \cos \theta -1)=a$$ $${\sin }^2\theta \cos \theta -\sin \theta +\sin \theta {\cos }^2\theta -\cos \theta =a$$ $$(1-{\cos }^2\theta )\cos \theta -\sin \theta +\sin \theta (1-{\sin }^2\theta )-\cos \theta =a$$ $$\cos \theta -{\cos }^3\theta -\sin \theta +\sin \theta -{\sin }^3\theta -\cos \theta =a$$ $$-{\cos }^3\theta -{\sin }^3\theta =a$$ $$\mathrm{for}\theta \in [0;-\frac{\pi }{2}]:$$ $$f\left(\theta \right)=-{\cos }^3\theta -{\sin }^3\theta \mathrm{has}\mathrm{got}\mathrm{a}\mathrm{local}\mathrm{maximum}$$ $$\mathrm{at}\left(\begin{array}{c}\frac{\pi }{4}\\ -\frac{\sqrt{2}}{2}\end{array}\right)\mathrm{and}\mathrm{local}\mathrm{mimima}\mathrm{at}\left(\begin{array}{c}0\\ -1\end{array}\right)\mathrm{and}\left(\begin{array}{c}\frac{\pi }{2}\\ -1\end{array}\right)$$ $$\Rightarrow \mathrm{for}a>-\frac{\sqrt{2}}{2}\mathrm{we}\mathrm{have}0\mathrm{solutions}$$ $$\mathrm{for}a=-\frac{\sqrt{2}}{2}\mathrm{we}\mathrm{have}1\mathrm{solution}$$ $$\mathrm{for}-1⩽a<-\frac{\sqrt{2}}{2}\mathrm{we}\mathrm{have}2\mathrm{solutions}$$ $$\mathrm{for}a<-1\mathrm{we}\mathrm{have}0\mathrm{solutions}$$

Commented bynaka3546 last updated on 05/May/19

$$syy=a?$$ $$Isitamistakeinwriting,sir?$$

Commented byMJS last updated on 05/May/19

$$\mathrm{sorry}\mathrm{just}\mathrm{a}\mathrm{weird}\mathrm{typo}.\mathrm{I}\mathrm{corrected}\mathrm{it}.$$

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