Trigonometrical Equations
normal

સમિકરણ $\frac{1}{2} +cosx + cos2x + cos3x + cos4x = 0$ નો ઉકેલ . . . . મેળવો.

A

$x=\frac{2n\pi}{9},n\in I,n\neq 9m,m\in I$

B

$x=\frac{2n\pi}{9},n\in I,n= 9m,m\in I$

C

$x=\frac{n\pi}{9}+\frac{\pi}{2},n\in I$

D

$x=\frac{2n\pi}{3}+\frac{\pi}{6},n\in I$

Solution

$\frac{1}{2}+\frac{1}{2 \sin \frac{x}{2}} 2 \sin \frac{x}{2}(\cos x+\cos 2 x+\cos 3 x+\cos 4 x)=0$

$=\frac{1}{2}+\frac{1}{2 \sin \frac{x}{2}}\left(\sin \frac{9 x}{2}-\sin \frac{x}{2}\right)=0$

$=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}=0 $ 

$ \Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{9}, \mathrm{n} \neq 9 \mathrm{mm} \in \mathrm{I}$

Standard 11
Mathematics

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