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प्रश्न
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
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उत्तर
\[\sin2x, \frac{1}{2} and \cos2x are in AP . \]
\[ \therefore \sin2x + \cos2x = 2 \times \frac{1}{2}\]
\[ \Rightarrow \sin2x + \cos2x = 1 . . . (1)\]
This equation is of the form \[a \sin\theta + b \cos\theta = c\], where
a = 1, b = 1 and c = 1
Now,
Let: \[a = r \sin \alpha\] and \[b = r \cos \alpha\]
Thus, we have:
\[r \sin \alpha \sin2x + r \cos\alpha \cos2x = 1\]
\[\Rightarrow r \cos (2x - \alpha) = 1\]
\[ \Rightarrow \sqrt{2} \cos \left( 2x - \frac{\pi}{4} \right) = 1\]
\[ \Rightarrow \cos \left( 2x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \cos \left( 2x - \frac{\pi}{4} \right) = \cos \frac{\pi}{4}\]
\[ \Rightarrow 2x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} , n \in Z\]
Taking positive value, we get:
\[ \Rightarrow 2x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}\]
\[ \Rightarrow x = n\pi + \frac{\pi}{4}\]
Taking negative value, we get:
\[ \Rightarrow 2x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}\]
\[ \Rightarrow 2x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}\]
\[ \Rightarrow x = n\pi, n \in Z\]
For n = 0, the values of x are \[\frac{\pi}{4} and 0\] and for n = 1, the values of x are `(5pi)/4` and π
For the other value of n, the given condition is not true, i.e., [0, π].
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