2sin-2x-4sin-2-x-7cos-2x-with-pi-2-lt-x-lt-pi-find-sin-2x-

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 by bemath 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

\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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