4sin^( 2) (x)− 4sin(x)=cos^( 2) (x)−2cos(x) , 0<x<(π/2) . Find sin(x)=?


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Question Number 173787 by mnjuly1970 last updated on 18/Jul/22

$4 {s i n}^{2} (x) - 4 s i n (x) = {c o s}^{2} (x) - 2 c o s (x)$ $, 0 < x < \frac{π}{2} . F i n d s i n (x) = ?$
Commented byinfinityaction last updated on 18/Jul/22

$\frac{1}{\sqrt{5}}$
Commented bysom(math1967) last updated on 18/Jul/22

$\frac{3}{5} a l s o$
Commented byinfinityaction last updated on 18/Jul/22

${4 \sin}^{2} x - 4 \sin x + 1 = \cos^{2} x - 2 \cos x + 1$ ${(1 - 2 \sin x)}^{2} = {(1 - \cos x)}^{2}$ ${(1 - 2 \sin x)}^{2} - {(1 - \cos x)}^{2} = 0$ $(\cos x - 2 \sin x) (2 - 2 \sin x - \cos x) = 0$ $i f \cos x - 2 \sin x = 0 \Rightarrow \tan x = \frac{1}{2}$ $\sin x = \frac{1}{\sqrt{5}}$ $2 - 2 \sin x = \sqrt{1 - \sin^{2} x}$ ${5 \sin}^{2} x - 8 \sin x + 3 = 0$ $(\sin x - 1) (5 \sin x - 3) = 0$ $\sin x = 1 (r e j e c t e d)$ $\sin x = \frac{3}{5}$
Commented bymnjuly1970 last updated on 18/Jul/22

$t h a n k y o u s o m u c h s i r . .$
Commented byMJS_new last updated on 18/Jul/22

$why do you reject 1 ?$
Commented byinfinityaction last updated on 18/Jul/22

$g i v e n i n t h e q u e s t i o n$ $0 < x < \frac{π}{2}$
Commented byTawa11 last updated on 18/Jul/22

$Great sirs$
Commented byMJS_new last updated on 18/Jul/22

$ok {didn}^{'} t notice . . .$
Commented byinfinityaction last updated on 18/Jul/22

$n o p r o b l e m s i r y o u r a l g e b r a$ $i s v e r y g o o d$ $i a m s t u d e n t o f u n d e r g r u a t e$ $s o c a n y o u s u g g e s t m e a n$ $a l g e b r a b o o k ? ?$
Commented byMJS_new last updated on 18/Jul/22

$sorry I {can}^{'} t . I studied math more than 20$ $years ago but I became a musician . math is$ $only my recreational sport now . . .$
Commented byinfinityaction last updated on 19/Jul/22

$o k s i r t h a n k s$
	Answered by MJS_new last updated on 18/Jul/22
	$x=2arctan t ((−8t^3 +16t^2 −8t)/((t^2 +1)^2 ))=((3t^4 −2t^2 −1)/((t^2 +1)^2 )) −8t(t−1)^2 =(t−1)(t+1)(3t^2 +1) t_1 =1 3t^3 +11t^2 −7t+1=0 (3t−1)(t+2−(√5))(t+2+(√5))=0 t_2 =(1/3) t_3 =−2+(√5) t_4 =−2−(√5) sin x =sin 2arctan t =((2t)/(t^2 +1)) sin x ∈{−((√5)/5), ((√5)/5), (3/5), 1}$
	$x = 2 \arctan t$ $\frac{- 8 t^{3} + 16 t^{2} - 8 t}{{(t^{2} + 1)}^{2}} = \frac{3 t^{4} - 2 t^{2} - 1}{{(t^{2} + 1)}^{2}}$ $- 8 t {(t - 1)}^{2} = (t - 1) (t + 1) (3 t^{2} + 1)$ $t_{1} = 1$ $3 t^{3} + 11 t^{2} - 7 t + 1 = 0$ $(3 t - 1) (t + 2 - \sqrt{5}) (t + 2 + \sqrt{5}) = 0$ $t_{2} = \frac{1}{3}$ $t_{3} = - 2 + \sqrt{5}$ $t_{4} = - 2 - \sqrt{5}$ $\sin x = \sin 2 \arctan t = \frac{2 t}{t^{2} + 1}$ $\sin x \in {- \frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}, \frac{3}{5}, 1}$
	Answered by a.lgnaoui last updated on 18/Jul/22

	$4 \sin^{2} (x) - 4 \sin (x) = 1 - \sin^{2} (x) - 2 \sqrt{1 - \sin^{2} (x)}$ $p o s i n s \sin (x) = z$ $5 z^{2} - 4 z - 1 = - 2 \sqrt{1 - z^{2}}$ ${[z - 1]}^{2} {[5 z + 1]}^{2} = 4 (1 - z^{2})$ $(1 - z) {(5 z + 1)}^{2} = 4 (1 + z)$ $(1 - z) (25 z^{2} + 10 z + 1) = 4 z + 4$ $25 z^{2} + 10 z + 1 - 25 z^{3} - 10 z^{2} - z - 4 z - 4 = 0$ $- 25 z^{3} + 15 z^{2} - 5 z - 3 = 0$ $25 z^{3} - 15 z^{2} + 5 z + 3 = 0$ $z^{3} - \frac{3}{5} z^{2} + \frac{z}{5} + \frac{3}{25} = 0$ $Z = z + 1 / 5$ $Z^{3} = z^{3} + \frac{3}{5} z^{2} + \frac{3}{25} z + \frac{1}{125}$ $- \frac{3}{5} Z^{2} = \frac{- 3}{5} z^{2} + - \frac{6}{25} z + \frac{- 3}{125}$ $Z = z + \frac{1}{5}$ $Z^{3} + (\frac{2}{25}) Z + \frac{22}{125} = 0 p = \frac{2}{25} q = \frac{22}{125}$ $p^{3} = (4 \times 8 / 5^{6}) / 27 + (4 \times 11^{2}) /$ $Δ = 484 / 5^{6} + \frac{32}{27 \times 5^{6}} = \frac{484 \times 27 + 32}{3^{3} \times 5^{6}} = \frac{13068 + 32}{} = \frac{13100}{421875} = 0, 031 \Rightarrow \sqrt{Δ} = 0, 176$ $q = \frac{22}{125} = 0, 176 \Rightarrow \frac{q}{2} = 0, 088$ $= - q / 2 \pm [^{3} \sqrt{\frac{- q}{2} \pm (4 p^{3} + 27 q^{2}) / 27)}] / 2 \sqrt{0, 176 - 0, 088}$ $Z = 0, 088 + [^{3} \sqrt{0, 088 + 0, 176}] / 2$ $Z = 0, 088 + [^{3} \sqrt{- 0, 088 + 0, 176}] / 2$ $Z = 0, 64 e t Z = - 0, 56$ $z = Z - \frac{1}{5} \Rightarrow \sin (x) = 0, 44$ $\sin (x) = 0, 36$
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