Question Number 18588 by Joel577 last updated on 25/Jul/17
$$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{a}\:+\:\mathrm{3}}{{a}\:+\:\mathrm{1}} \\ $$$$\mathrm{How}\:\mathrm{many}\:{a}\:\mathrm{that}\:\mathrm{can}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{above}? \\ $$
Answered by ajfour last updated on 25/Jul/17
![−1≤((2a+3)/(a+1))≤1 ; a≠−1 CaseI: if a>−1 −a−1≤2a+3 and 2a+3≤a+1 ⇒ a≥−(4/3) and a≤−2 both cannot be true simultaneously Case II: a<−1 −a−1≥2a+3 and 2a+3≥a+1 ⇒ a≤−(4/3) and a≥−2 with a<−1 ⇒ a ∈ [−2, −(4/3)] .](https://www.tinkutara.com/question/Q18599.png)
$$−\mathrm{1}\leqslant\frac{\mathrm{2a}+\mathrm{3}}{\mathrm{a}+\mathrm{1}}\leqslant\mathrm{1}\:\:\:;\:\:\mathrm{a}\neq−\mathrm{1} \\ $$$$\mathrm{CaseI}:\:\:\mathrm{if}\:\mathrm{a}>−\mathrm{1} \\ $$$$−\mathrm{a}−\mathrm{1}\leqslant\mathrm{2a}+\mathrm{3}\:\:\mathrm{and}\:\:\mathrm{2a}+\mathrm{3}\leqslant\mathrm{a}+\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{a}\geqslant−\frac{\mathrm{4}}{\mathrm{3}}\:\:\mathrm{and}\:\:\mathrm{a}\leqslant−\mathrm{2} \\ $$$$\mathrm{both}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}\:\mathrm{simultaneously} \\ $$$$\mathrm{Case}\:\mathrm{II}:\:\:\:\mathrm{a}<−\mathrm{1} \\ $$$$−\mathrm{a}−\mathrm{1}\geqslant\mathrm{2a}+\mathrm{3}\:\:\:\:\mathrm{and}\:\:\mathrm{2a}+\mathrm{3}\geqslant\mathrm{a}+\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\mathrm{a}\leqslant−\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\mathrm{and}\:\:\mathrm{a}\geqslant−\mathrm{2} \\ $$$$\mathrm{with}\:\:\mathrm{a}<−\mathrm{1}\: \\ $$$$\:\Rightarrow\:\:\:\:\:\:\mathrm{a}\:\in\:\left[−\mathrm{2},\:−\frac{\mathrm{4}}{\mathrm{3}}\right]\:. \\ $$
Commented by Joel577 last updated on 25/Jul/17
$${thank}\:{you}\:{very}\:{much} \\ $$