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Question Number 22729 by selestian last updated on 22/Oct/17

Commented by math solver last updated on 22/Oct/17

$0$
Commented by selestian last updated on 22/Oct/17

$s i r p l e a s e s e n d m e t h e s o l u t i o n$
Commented by math solver last updated on 22/Oct/17

$i h a v e s e n t . s e e b e l o w$
Commented by math solver last updated on 22/Oct/17

$h e y, i {d o n}^{'} t k n o w i f y o u h a v e s t u d i e d$ $a m, g m .$ ${d o n}^{'} t w o r r y, i h a v e a n o t h e r s o l . t o o :)$
Commented by selestian last updated on 22/Oct/17

$y e s i h a d s t u d i e d a m n d g m t h a n k s$
	Answered by math solver last updated on 22/Oct/17

	Answered by $@ty@m last updated on 24/Oct/17
	$Let 2^x =y ⇒sin (e^x )=y+(1/y) ⇒y^2 −sin (e^x )y+1=0 −−−(1) D=b^2 −4ac =sin^2 (e^x )−4 ⇒0−4≤D≤1−4 {∵0≤sin^2 θ≤1} ⇒−4≤D≤−3 ⇒ (1) has no real root.$
	$L e t 2^{x} = y$ $\Rightarrow \sin (e^{x}) = y + \frac{1}{y}$ $\Rightarrow y^{2} - \sin (e^{x}) y + 1 = 0 - - - (1)$ $D = b^{2} - 4 a c$ $= \sin^{2} (e^{x}) - 4$ $\Rightarrow 0 - 4 ⩽ D ⩽ 1 - 4$ ${∵ 0 ⩽ \sin^{2} θ ⩽ 1}$ $\Rightarrow - 4 ⩽ D ⩽ - 3$ $\Rightarrow (1) h a s n o r e a l r o o t .$
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