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Trigonometrical Equations
hard
$\sin x - 3\sin 2x + \sin 3x = $ $\cos x - 3\cos 2x + \cos 3x$ का व्यापक हल है
A
$n\pi + \frac{\pi }{8}$
B
$\frac{{n\pi }}{2} + \frac{\pi }{8}$
C
${( - 1)^n}\frac{{n\pi }}{2} + \frac{\pi }{8}$
D
$2n\pi + {\cos ^{ - 1}}\frac{3}{2}$
(IIT-1989)
Solution
(b) $\sin x – 3\sin 2x + \sin 3x = \cos x – 3\cos 2x + \cos 3x$
$ \Rightarrow $$2\sin 2x\cos x – 3\sin 2x – 2\cos 2x\cos x + 3\cos 2x = 0$
$ \Rightarrow $ $\sin 2x(2\cos x – 3) – \cos 2x(2\cos x – 3) = 0$
$ \Rightarrow $$(\sin 2x – \cos 2x)\,\,(2\cos x – 3) = 0 \Rightarrow \sin 2x = \cos 2x$
$(\,\because \cos x \ne 3/2)$
$ \Rightarrow $ $2x = 2n\pi \pm \left( {\frac{\pi }{2} – 2x} \right)$
अर्थात् $x = \frac{{n\pi }}{2} + \frac{\pi }{8}$.
Standard 11
Mathematics
