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The number of solutions of $sin \,3x\, = cos\, 2x$ , in the interval $\left( {\frac{\pi }{2},\pi } \right)$ is
$3$
$4$
$2$
$1$
Solution
Given $\sin \,3x\, = \,\cos 2x$
$ \Rightarrow {\kern 1pt} \sin \,3x\, = \,\sin \,(\pi /2 – 2x)$
We now that $\sin A = \sin B$
$ \Rightarrow \,A\, = \,n\pi + {( – )^n}B,$ where $n$ is an integer
Using the above identity , we get
$3x = n\pi \, + \,{( – 1)^n}(\pi /2 – 2x)$
$n = 1,\,\,x = \pi – \pi /2\, \Rightarrow \,x\, = \,\pi /2 \notin \,(\pi /2,\pi )$
$n = 2,\,\,x = \pi /2\, \notin \,\,(\pi /2,\pi )$
$n = 3,\,\,x = 5\pi /2\, \notin \,\,(\pi /2,\pi )$
$n = 4,\,\,x = 9\pi /10\, \notin \,\,(\pi /2,\pi )$
$n = 5,\,\,x = 9\pi /2 = \pi \, \notin \,\,(\pi /2,\pi )$
Now, for all negative integers $x$ would be negative.
For all value of $n > 5$ , solution $> \pi $
Hence the only possible solution is for $n=4$
and $x=9\pi /10$.
