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प्रश्न
Find the general solution of the following equation:
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उत्तर
We have:
\[\Rightarrow \cos x = - \sin 2x\]
\[ \Rightarrow \cos x = \cos \left( \frac{\pi}{2} + 2x \right)\]
\[ \Rightarrow x = 2n\pi \pm \left( \frac{\pi}{2} + 2x \right), n \in Z\]
On taking positive sign, we have:
\[x = 2n\pi + \frac{\pi}{2} + 2x\]
\[ \Rightarrow - x = 2n\pi + \frac{\pi}{2}\]
\[ \Rightarrow x = 2m\pi - \frac{\pi}{2}, m = - n \in Z\]
\[ \Rightarrow x = \frac{(4m - 1)\pi}{2}, m \in Z\]
On taking negative sign, we have:
`x-2nx-x/2-2x`
`=>3x=2nx-pi/2`
`=>x=((4n-1)x)/6,n in "Z"`
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