1−2sin(4x)<cos^2 (4x) solve.


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Question Number 57357 by ANTARES VY last updated on 03/Apr/19

$1 - 2 \sin (4 x) < \cos^{2} (4 x)$ $solve .$
Commented by bshahid010@gmail.com last updated on 03/Apr/19

$1 - 2 \sin (4 x) < \cos^{2} (4 x)$ $1 - 2 \sin (4 x) - \cos^{2} (4 x) < 0$ $1 - 2 \sin (4 x) - 1 + \sin^{2} (4 x) < 0$ $\sin^{2} 4 x - 2 \sin 4 x < 0$ $\sin 4 x (\sin 4 x - 2) < 0$ $\sin 4 x \in (0, 2) so \sin 4 x \in (0, 1)$ $x \in (0, \sin 1)$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

$m a x v a l u e o f s i n θ = 1$ $s o i t h i n k s i n 4 x \in (0, 1) b u t$ $s i n 4 x \notin (0, 2)$ $l e t o t h e r c o m m e n t . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

	$s i n 4 x = a$ $1 - 2 a < 1 - a^{2}$ $a^{2} - 2 a < 0$ $a (a - 2) < 0$ $a \neq 2 a_{m a x} = 1$ $1 > a > 0 i s t h e r e a u i r e d s o l u t i o n$ $1 > s i n 4 x > 0$ $s i n \frac{π}{2} > s i n 4 x > s i n 0$ $\frac{π}{8} > x > 0$
	Commented by ANTARES VY last updated on 03/Apr/19

	$? ? ?$
	Answered by einsteindrmaths@hotmail.fr last updated on 03/Apr/19
	$cos^2 (4x)=1−sin^2 (4x)====>1−sin(4x)<1−sin^2 (4x) ==>sin^2 (4x)−sin(4x)<0==>sin(4x)(1−sin(4x))<0==> sin(4x)>0&1−sin(4x)≠0===>x≠{(π/8)+((kπ)/2)/k∈IZ}∪]((kπ)/(2 ));(((2k+1)π)/4)[$
	$c os^{2} (4 x) = 1 - {s i n}^{2} (4 x) ====> 1 - \sin (4 x) < 1 - {s i n}^{2} (4 x)$ $==> {s i n}^{2} (4 x) - s i n (4 x) < 0 ==> s i n (4 x) (1 - s i n (4 x)) < 0 ==>$ $s i n (4 x) > 0 & 1 - s i n (4 x) \neq 0 ===> x \neq {\frac{π}{8} + \frac{k π}{2} / k \in I Z} \cup] \frac{k π}{2}; \frac{(2 k + 1) π}{4} [$
	Answered by mr W last updated on 03/Apr/19
	$1−2 sin (4x)<1−sin^2 (4x) sin (4x){sin (4x)−2}<0 since sin (4x)−2<0 ⇒sin (4x)>0 or ⇒0<sin (4x)≤1 ⇒4x∈(2nπ,2nπ+π) ⇒x∈(((nπ)/2),((nπ)/2)+(π/4)) with n∈W$
	$1 - 2 \sin (4 x) < 1 - \sin^{2} (4 x)$ $\sin (4 x) {\sin (4 x) - 2} < 0$ $s i n c e \sin (4 x) - 2 < 0$ $\Rightarrow \sin (4 x) > 0$ $o r$ $\Rightarrow 0 < \sin (4 x) ⩽ 1$ $\Rightarrow 4 x \in (2 n π, 2 n π + π)$ $\Rightarrow x \in (\frac{n π}{2}, \frac{n π}{2} + \frac{π}{4}) w i t h n \in W$
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