The number of solutions of sin ,3x, = cos, 2x | vedclass

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Question

Given $\sin \,3x\, = \,\cos 2x$

$ \Rightarrow {\kern 1pt} \sin \,3x\, = \,\sin \,(\pi /2 - 2x)$

We now that $\sin A = \sin B$

$ \Ri

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$1$

Vedclass · Answer

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

$3x = n\pi \, + \,{( - 1)^n}(\pi /2 - 2x)$

$n = 1,\,\,x = \pi  - \pi /2\, \Rightarrow \,x\, = \,\pi /2 
otin \,(\pi /2,\pi )$

$n = 2,\,\,x = \pi /2\, 
otin \,\,(\pi /2,\pi )$

$n = 3,\,\,x = 5\pi /2\, 
otin \,\,(\pi /2,\pi )$

$n = 4,\,\,x = 9\pi /10\, 
otin \,\,(\pi /2,\pi )$

$n = 5,\,\,x = 9\pi /2 = \pi \, 
otin \,\,(\pi /2,\pi )$

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 $x=9\pi /10$.

The number of solutions of $sin \,3x\, = cos\, 2x$ , in the interval $\left( {\frac{\pi }{2},\pi } \right)$ is

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