2sin 2x −4sin^2 x = 7cos 2x with (π/2)< x


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Question Number 104576 by bemath last updated on 22/Jul/20

$2 \sin 2 x - 4 \sin^{2} x = 7 \cos 2 x$ $w i t h \frac{π}{2} < x < π$ $f i n d \sin 2 x$
Commented bybemath last updated on 22/Jul/20

$t h a n k y o u b o t h$
	Answered by bobhans last updated on 22/Jul/20
	$⇔4sin x cos x−4sin^2 x = 7(cos^2 x−sin^2 x) 4sin x (cos x−sin x) = 7(cos x−sin x)(cos x+sin x) ⇒(sin x−cos x){4sin x−7cos x−7sin x)=0 { ((sin x=cos x (rejected))),((7cos x = −3sin x )) :} tan x = −(7/3) → { ((sin x = (7/(√(58))))),((cos x = −(3/(√(58))))) :} sin 2x = 2×(7/(√(58))) ×−(3/(√(58))) = −((21)/(29)) ★$
	$\Leftrightarrow 4 \sin x \cos x - 4 \sin^{2} x = 7 (\cos^{2} x - \sin^{2} x)$ $4 \sin x (\cos x - \sin x) = 7 (\cos x - \sin x) (\cos x + \sin x)$ $\Rightarrow (\sin x - \cos x) {4 \sin x - 7 \cos x - 7 \sin x) = 0$ ${\begin{cases} \sin x = \cos x (r e j e c t e d) \\ 7 \cos x = - 3 \sin x \end{cases}$ $\tan x = - \frac{7}{3} \to {\begin{cases} \sin x = \frac{7}{\sqrt{58}} \\ \cos x = - \frac{3}{\sqrt{58}} \end{cases}$ $\sin 2 x = 2 \times \frac{7}{\sqrt{58}} \times - \frac{3}{\sqrt{58}} = - \frac{21}{29} ★$
	Answered by OlafThorendsen last updated on 22/Jul/20

	$2 \sin 2 x - 4 \frac{1 - \cos 2 x}{2} = 7 \cos 2 x$ $2 \sin 2 x - 2 = 5 \cos 2 x$ $\Rightarrow {4 \sin}^{2} 2 x - 8 \sin 2 x + 4 = {25 \cos}^{2} 2 x$ ${4 \sin}^{2} 2 x - 8 \sin 2 x + 4 = 25 - {25 \sin}^{2} 2 x$ ${29 \sin}^{2} x - 8 \sin 2 x - 21 = 0$ $(\sin 2 x - 1) (29 \sin 2 x + 21) = 0$ $\sin 2 x = 1 \Leftrightarrow x = \frac{π}{4} + k π, k \in Z$ $impossible, \frac{π}{2} < x < π$ $\sin 2 x = - \frac{21}{29} is a solution .$
	Answered by Dwaipayan Shikari last updated on 22/Jul/20

	$2 s i n 2 x - 4 {s i n}^{2} x = 7 {c o s}^{2} x - 7 {s i n}^{2} x$ $- 7 {c o s}^{2} x + 4 s i n x c o s x + 3 {s i n}^{2} x = 0$ $- 7 {c o s}^{2} x + 7 s i n x c o s x - 3 s i n x c o s x + 3 {s i n}^{2} x = 0$ $- 7 c o s x (c o s x - s i n x) - 3 s i n x (c o s x - s i n x) = 0$ $(c o s x - s i n x) (7 c o s x + 3 s i n x) = 0$ $c o s x = s i n x$ $t a n x = 1$ $\frac{2 t a n x}{{t a n}^{2} x + 1} = s i n 2 x = 1 (B u t i t i s n o t t h e s o l u t i o n)$ $7 c o s x = - 3 s i n x$ $t a n x = \frac{- 7}{3}$ $\frac{2 t a n x}{{t a n}^{2} x + 1} = \frac{\frac{- 14}{3}}{\frac{49}{9} + 1} = \frac{- 14.3}{58} = - \frac{21}{29}$
	Answered by ajfour last updated on 22/Jul/20

	$l e t \cos 2 x = t$ $\Rightarrow 4 (1 - t^{2}) = {(7 t + 2 - 2 t)}^{2}$ $4 - 4 t^{2} = 25 t^{2} + 20 t + 4$ $\Rightarrow 29 t^{2} + 20 t = 0$ $\Rightarrow t = - \frac{20}{29}, \sin 2 x = - \sqrt{1 - t^{2}}$ $\sin 2 x = - \frac{\sqrt{49 \times 9}}{29} = - \frac{21}{29} .$
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