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

Question Number 147581 by EDWIN88 last updated on 22/Jul/21

2 \sin 2 x - 4 \sin^{2} x = 7 \cos 2 x

\frac{π}{2} < x < π \Rightarrow \sin 2 x = ?

Answered by iloveisrael last updated on 22/Jul/21

If 2 \sin 2 x - 4 \sin^{2} x = 7 \cos 2 x

where \frac{π}{2} < x < π

Then \sin 2 x = ?

π < x < 2 π \Rightarrow \sin 2 x < 0

2 \sin 2 x - 4 (\frac{1 - \cos 2 x}{2}) = 7 \cos 2 x

2 \sin 2 x - 2 + 2 \cos 2 x = 7 \cos 2 x

2 \sin 2 x - 5 \cos 2 x - 2 = 0

2 \tan 2 x - 5 = 2 \sec 2 x

4 \tan^{2} 2 x - 20 \tan 2 x + 25 = 4 + 4 \tan^{2} 2 x

20 \tan 2 x = 21

\tan 2 x = \frac{21}{20} \Rightarrow \sin 2 x = - \frac{21}{\sqrt{21^{2} + 20^{2}}}

\sin 2 x = - \frac{20}{\sqrt{841}} = - \frac{20}{29} .