Question Number 59113 by naka3546 last updated on 04/May/19
$${Let}\:\:{a}\:\:{is}\:\:{a}\:\:{real}\:\:{number}\:.\:\:{How}\:\:{many}\:\:{solutions} \\ $$$${can}\:\:{the}\:\:{equation}\:\:{in}\:\:\theta\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:−\:\mathrm{1}\right)\:\:=\:\:{a} \\ $$$${have}\:\:{for}\:\:\mathrm{0}\:<\:\theta\:<\:\frac{\pi}{\mathrm{2}}\:\:? \\ $$
Answered by MJS last updated on 05/May/19
![(sin θ +cos θ)(sin θ cos θ −1)=a sin^2 θ cos θ −sin θ +sin θ cos^2 θ −cos θ =a (1−cos^2 θ)cos θ −sin θ +sin θ (1−sin^2 θ)−cos θ =a cos θ −cos^3 θ −sin θ +sin θ −sin^3 θ −cos θ =a −cos^3 θ −sin^3 θ =a for θ∈[0; −(π/2)]: f(θ)=−cos^3 θ −sin^3 θ has got a local maximum at (((π/4)),((−((√2)/2))) ) and local mimima at ((0),((−1)) ) and (((π/2)),((−1)) ) ⇒ for a>−((√2)/2) we have 0 solutions for a=−((√2)/2) we have 1 solution for −1≤a<−((√2)/2) we have 2 solutions for a<−1 we have 0 solutions](https://www.tinkutara.com/question/Q59119.png)
$$\left(\mathrm{sin}\:\theta\:+\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:−\mathrm{1}\right)={a} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta\:+\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{cos}\:\theta\:={a} \\ $$$$\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta\:+\mathrm{sin}\:\theta\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)−\mathrm{cos}\:\theta\:={a} \\ $$$$\mathrm{cos}\:\theta\:−\mathrm{cos}^{\mathrm{3}} \:\theta\:−\mathrm{sin}\:\theta\:+\mathrm{sin}\:\theta\:−\mathrm{sin}^{\mathrm{3}} \:\theta\:−\mathrm{cos}\:\theta\:={a} \\ $$$$−\mathrm{cos}^{\mathrm{3}} \:\theta\:−\mathrm{sin}^{\mathrm{3}} \:\theta\:={a} \\ $$$$\mathrm{for}\:\theta\in\left[\mathrm{0};\:−\frac{\pi}{\mathrm{2}}\right]: \\ $$$${f}\left(\theta\right)=−\mathrm{cos}^{\mathrm{3}} \:\theta\:−\mathrm{sin}^{\mathrm{3}} \:\theta\:\mathrm{has}\:\mathrm{got}\:\mathrm{a}\:\mathrm{local}\:\mathrm{maximum} \\ $$$$\mathrm{at}\:\begin{pmatrix}{\frac{\pi}{\mathrm{4}}}\\{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix}\:\:\mathrm{and}\:\mathrm{local}\:\mathrm{mimima}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{1}}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{\frac{\pi}{\mathrm{2}}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\:\mathrm{for}\:{a}>−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{have}\:\mathrm{0}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:\:\mathrm{for}\:{a}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}\:\mathrm{solution} \\ $$$$\:\:\:\:\:\:\mathrm{for}\:−\mathrm{1}\leqslant{a}<−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:\:\mathrm{for}\:{a}<−\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\mathrm{0}\:\mathrm{solutions} \\ $$
Commented by naka3546 last updated on 05/May/19
$${syy}\:\:=\:\:{a}\:\:? \\ $$$${Is}\:\:{it}\:\:{a}\:\:{mistake}\:\:{in}\:\:{writing},\:\:{sir}\:\:? \\ $$
Commented by MJS last updated on 05/May/19
$$\mathrm{sorry}\:\mathrm{just}\:\mathrm{a}\:\mathrm{weird}\:\mathrm{typo}.\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it}. \\ $$