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

(i)
$$\begin{gathered} \sin 2x,\frac{1}{2}\mathrm{and}\cos 2x\mathrm{are}\mathrm{in}\mathrm{AP}. \\ \therefore \sin 2x+\cos 2x=2\times \frac{1}{2} \\ \Rightarrow \sin 2x+\cos 2x=1...\left(1\right) \end{gathered}$$

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 $\tan \alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (1), we get:
$r\sin \alpha \sin 2x+r\cos \alpha \cos 2x=1$
$$\begin{gathered} \Rightarrow r\cos (2x-\alpha )=1 \\ \Rightarrow \sqrt{2}\cos \left(2x-\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(2x-\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \left(2x-\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z} \\ \mathrm{Taking}\mathrm{positive}\mathrm{value},\mathrm{we}\mathrm{get}: \\ \Rightarrow 2\mathrm{x}-\frac{\mathrm{\pi }}{4}=2\mathrm{n}\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \mathrm{Taking}\mathrm{negative}\mathrm{value},\mathrm{we}\mathrm{get}: \\ \Rightarrow 2\mathrm{x}-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ \Rightarrow 2\mathrm{x}-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ \Rightarrow x=n\mathrm{\pi },n\in Z \end{gathered}$$

For n = 0, the values of x are $\frac{\mathrm{\pi }}{4}\mathrm{and}0$ and for n = 1, the values of x are $\frac{5\mathrm{\pi }}{4}\mathrm{and}\mathrm{\pi }$.
$\frac{5\mathrm{\pi }}{4}\mathrm{does}\mathrm{not}\mathrm{satisfy}\mathrm{the}\mathrm{condition}.$
For the other value of n, the given condition is not true, i.e., [0, π].