Question

Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]

 and cos 2x are in A.P.

Answer 1

\[\sin2x, \frac{1}{2} and \cos2x are in AP . \]
\[ \therefore \sin2x + \cos2x = 2 \times \frac{1}{2}\]
\[ \Rightarrow \sin2x + \cos2x = 1 . . . (1)\]
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:

\[r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\] and `tanalpha =1=>alpha=pi/4`

On putting \[a = 1 = r \sin \alpha\] and \[b = 1 = r \cos \alpha\] in equation (1), we get:
\[r \sin \alpha \sin2x + r \cos\alpha \cos2x = 1\]

\[\Rightarrow r \cos (2x - \alpha) = 1\]

\[ \Rightarrow \sqrt{2} \cos \left( 2x - \frac{\pi}{4} \right) = 1\]

\[ \Rightarrow \cos \left( 2x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \cos \left( 2x - \frac{\pi}{4} \right) = \cos \frac{\pi}{4}\]

\[ \Rightarrow 2x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} , n \in Z\]

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

\[\frac{5\pi}{4} \text{ does not satisfy the condition.}\]

For the other value of n, the given condition is not true, i.e., [0, π].