Trigonometrical Equations
medium

यदि $\cos 3x + \sin \left( {2x - \frac{{7\pi }}{6}} \right) =  - 2$, तब $x = $ (जहाँ $k \in Z$)

A

$\frac{\pi }{3}(6k + 1)$

B

$\frac{\pi }{3}(6k - 1)$

C

$\frac{\pi }{3}(2k + 1)$

D

इनमें से कोई नहीं

Solution

हमें ज्ञात है कि, $\cos 3x + \sin {\rm{ }}\left( {2x – \frac{{7\pi }}{6}} \right)\, =  – 2$

$ \Rightarrow $ $1 + \cos 3x + 1 + \sin \left( {2x – \frac{{7\pi }}{6}} \right) = 0$

$ \Rightarrow $ $(1 + \cos 3x) + 1 – \cos \left( {2x – \frac{{2\pi }}{3}} \right) = 0$

$ \Rightarrow $ $2{\cos ^2}\frac{{3x}}{2} + 2{\sin ^2}\left( {x – \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $ $\cos \frac{{3x}}{2} = 0$ तथा $\sin \left( {x – \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $ $\frac{{3x}}{2} = \frac{\pi }{2},\,\frac{{3\pi }}{2},\,…..$

तथा $x – \frac{\pi }{3}$=0, $\pi ,2\pi ….. \Rightarrow x = \frac{\pi }{3}$

अत: $\cos \frac{{3x}}{2} = 0$ और $\sin \left( {x – \frac{\pi }{3}} \right) = 0$ का व्यापक हल

$x = 2k\pi  + \frac{\pi }{3} = \frac{\pi }{3}(6k + 1)$ है, जहाँ $k \in Z$.

Standard 11
Mathematics

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