Find the value of a for which the equation sin^4 x+asin^2 x+1=0 will have a solution.


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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

$p l s e x p l a i n t h i s :$ $H e n c e, t^{2} + a t + 1 = 0 s h o u l d h a v e a t l e a s t$ $o n e s o l u t i o n i n [0, 1] . . . . . . . . . . m u s t h a v e$ $e x a c t l y o n e r o o t i n [0, 1] .$
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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