Find-the-value-of-a-for-which-the-equation-sin-4-x-asin-2-x-1-0-will-have-a-solution-

Question Number 32220 by rahul 19 last updated on 21/Mar/18

F i n d t h e v a l u e o f a f o r w h i c h t h e e q u a t i o n

\sin^{4} x + a \sin^{2} x + 1 = 0 w i l l h a v e a s o l u t i o n .

Commented by rahul 19 last updated on 21/Mar/18

Commented by rahul 19 last updated on 21/Mar/18

p l s e x p l a i n t h i s :

H e n c e, t^{2} + a t + 1 = 0 s h o u l d h a v e a t l e a s t

o n e s o l u t i o n i n [0, 1] \dots \dots \dots . m u s t h a v e

e x a c t l y o n e r o o t i n [0, 1] .

Commented by MJS last updated on 21/Mar/18

because t = \sin^{2} x it should be

clear that t \in [0; 1] \Rightarrow t ≧ 0

t^{2} + a t + 1 = 0

(t - t_{1}) (t - t_{2}) = 0

t^{2} + (- t_{1} - t_{2}) t + t_{1} t_{2} = 0

\Rightarrow a = - t_{1} - t_{2} (but we {don}^{'} t need this)

1 = t_{1} t_{2} \land t ≧ 0 (from above) \Rightarrow

\Rightarrow t_{1} = t_{2}^{- 1} (but we {don}^{'} t need this)

so if {there}^{'} s one root in [0; 1],

{there}^{'} s also a change of sign, so

(f (0) < 0 \land f (1) > 0) \lor (f (0) > 0 \land f (1) < 0) \Rightarrow

\Rightarrow f (0) \times f (1) < 0

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