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Q.
The number of solutions of $[\sin x+\cos x]=3+[-\sin x]+[-\cos x]$ (where $[\cdot]$ denotes the greatest integer function), $x \in[0,2 \pi]$, is
Trigonometric Functions
Solution:
$[\sin x+\cos x]=3+[-\sin x]+[-\cos x]=1$
The maximum value of the L.H.S. is 1 and the minimum of the R.H.S. is also 1. Thus,
${[\sin x+\cos x]=3+[-\sin x]+[-\cos x]=1} $
or $[\sin x+\cos x]=1,[-\sin x]=-1,[-\cos x]=-1 $
$[\sin x+\cos x]=1 $
$\therefore 1 \leq \sin x+\cos x \leq \sqrt{2} $
$\therefore x \in\left[0, \frac{\pi}{2}\right]$...(i)
$\therefore {[-\sin x]=-1} $
$ \therefore -1 \leq-\sin x<0$
$\therefore 0<\cos x \leq 1 $
$ \therefore x \in\left(0, \frac{\pi}{2}\right] $ .... (ii)
$\therefore {[-\cos x]=-1} $
$ \therefore -1 \leq-\cos x < 0 $
$\therefore 0 < \cos x \leq 1 $
$ \therefore x \in\left(0, \frac{\pi}{2}\right]$ ....(iii)
From Eqs. (i), (ii), (iii), $x \in\left(0, \frac{\pi}{2}\right)$