- Home
- Standard 11
- Mathematics
Trigonometrical Equations
hard
यदि $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)\, + 2\cos x = 0$, तो
A
$x = \frac{\pi }{6}(4n + 1)$ या $x = \frac{\pi }{2}(4n - 1)$
B
$x = \frac{\pi }{6}(4n - 1)$ या $x = \frac{\pi }{2}(4n - 1)$
C
$x = \frac{\pi }{6}(4n + 1)$ या $x = \frac{\pi }{2}(4n + 1)$
D
इनमें से कोई नहीं
Solution
(a) $2(\sin x – \cos 2x) – \sin 2x(1 + 2\sin x) + 2\cos x = 0$
==> $2\sin x – 2 + 4{\sin ^2}x – 2\sin x\cos x – 4{\sin ^2}x\cos x$
$ + 2\cos x = 0$
==>$4{\sin ^2}x + 2\sin x – 2 – \cos x[4{\sin ^2}x + 2\sin x – 2]$= 0
==> $(1 – \cos x)\,(\sin x + 1)\,(4\sin x – 2) = 0$
अत: $\sin x = – 1$ या $\cos x = 1$ या $\sin x = \frac{1}{2}$
==> $x = (4n – 1)\,\frac{\pi }{2}$ तथा $x = (4n + 1)\frac{\pi }{6}$.
Standard 11
Mathematics
