The set of values of ′a′ for which all the solutions of the equation 4sin^4 x+asin^2 x+3=0 are real and distinct is ?


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Question Number 32563 by rahul 19 last updated on 27/Mar/18

$T h e s e t o f v a l u e s o f^{'} a^{'} f o r w h i c h$ $a l l t h e s o l u t i o n s o f t h e e q u a t i o n$ $4 \sin^{4} x + a \sin^{2} x + 3 = 0 a r e r e a l a n d$ $d i s t i n c t i s ?$
Commented by rahul 19 last updated on 27/Mar/18

$i t r i e d d o i n g t h e s a m e p r o c e d u r e$ $a s i n Q u e s . N o . 32220 .$
Commented by rahul 19 last updated on 27/Mar/18

$C o r r e c t a n s . i s [- 7, - 4 \sqrt{3}) .$
	Answered by rahul 19 last updated on 28/Mar/18

	$D ⩾ 0$ $a^{2} ⩾ 48$ $\Rightarrow a ⩾ 4 \sqrt{3} a n d a ⩽ - 4 \sqrt{3}$ $N o w l e t {s i n}^{2} x = t$ $\Rightarrow 4 t^{2} + a t + 3 = 0$ $\Rightarrow f (0) . f (1) < 0$ $\Rightarrow 3 \times (7 + a) < 0$ $\Rightarrow a < - 7 .$ $W {h a t}^{'} s w r o n g w i t h m y s o l u t i o n ?$
	Commented by rahul 19 last updated on 29/Mar/18

	$w h a t d o e s \lor, \land m e a n s i n y o u r$ $s o l u t i o n ?$
	Commented by MJS last updated on 29/Mar/18

	$I found it was off topic$ $look below$
	Commented by MJS last updated on 29/Mar/18

	$\land = ‘ ‘ {a n d}^{″}$ $\lor = ‘ ‘ {o r}^{″}$
	Answered by MJS last updated on 28/Mar/18

	$s^{2} + \frac{a}{4} s + \frac{3}{4} = 0$ $exactly 1 solution :$ ${(s - s_{0})}^{2} = s^{2} - 2 s_{0} s + s_{0}^{2}$ $- 2 s_{0} = \frac{a}{4} \land s_{0}^{2} = \frac{3}{4}$ $(a = 4 \sqrt{3} \land s = - \frac{\sqrt{3}}{2}) \lor (a = - 4 \sqrt{3} \land s = \frac{\sqrt{3}}{2})$ $BUT :$ $t^{4} + \frac{a}{4} t^{2} + \frac{3}{4} = 0 \Rightarrow t = \sqrt{s} \Rightarrow s ⩾ 0$ $\Rightarrow a = - 4 \sqrt{3} \land s = \frac{\sqrt{3}}{2} \land t = \pm \frac{\sqrt[4]{3} \sqrt{2}}{2}$ $this shows that the equation$ $f (x) = {4 \sin}^{4} x - 4 \sqrt{3} \sin^{2} x + 3 has$ $exactly 1 solution in [0; \frac{π}{2}]$ $with a > - 4 \sqrt{3} it has none$ $with - 7 ⩽ a < - 4 \sqrt{3} it has 2$ $with a < - 7 it has 1 again$ $a \to - \infty \Rightarrow x_{0} \to 0 but f (0) = 3 \forall a \in R$
	Commented by rahul 19 last updated on 28/Mar/18

	$s i r p l s c h e c k m y m e t h o d .$
	Commented by MJS last updated on 29/Mar/18

	$your method is ok but the$ $conclusions might be wrong$ $in qu . 32220 we were looking$ $for solutions, here {we}^{'} re looking$ $for all distinct solutions, so there$ $must be 4 in [0; π]$ $it seems difficult to understand$ $the nature of the connection$ $between the functions$ $f (x) = c_{1} \sin^{4} x + c_{2} \sin^{2} x + c_{3}$ $g (s) = c_{1} s^{4} + c_{2} s^{2} + c_{3}$ $h (t) = c_{1} t^{2} + c_{2} t + c_{3}$ $you should draw them or plot$ $them, if possible .$
	Commented by MJS last updated on 29/Mar/18

	Commented by MJS last updated on 29/Mar/18

	Commented by MJS last updated on 29/Mar/18

	Commented by MJS last updated on 29/Mar/18

	${we}^{'} re losing important information :$ $1 . picture$ ${4 \sin}^{4} x - 4 \sqrt{3} \sin^{2} x + 3$ ${4 \sin}^{4} x - {7 \sin}^{2} x + 3$ $[\frac{π}{4}; \frac{3 π}{4}]$ $2 . picture$ $4 s^{4} - 4 \sqrt{3} s^{2} + 3$ $4 s^{4} - 7 s^{2} + 3$ $[\frac{\sqrt{2}}{2}; 1 . 11^{()}]$ $3 . picture$ $4 t^{2} - 4 \sqrt{3} t + 3$ $4 t^{2} - 7 t + 3$ $[\frac{\sqrt{2}}{2}; 1 . 02^{()}]$ $(*) upper borders chosen for$ $symmetry of pics$
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