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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$1$

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

Similar Questions

If $\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3 ,$ then

If $\cos \theta + \cos 2\theta + \cos 3\theta = 0$, then the general value of $\theta $ is

Number of solution $(s)$ of the equation ${\cos ^2}2x + {\cos ^2}\frac{{5x}}{4} = \cos 2x\,{\cos ^2}5x$ in $\left[ {0,\frac{\pi }{3}} \right]$ is

If $m$ and $n$ respectively are the numbers of positive and negative value of $\theta$ in the interval $[-\pi, \pi]$ that satisfy the equation $\cos 2 \theta \cos \frac{\theta}{2}=\cos 3 \theta \cos \frac{9 \theta}{2}$, then $mn$ is equal to $.............$.

$\sin 6\theta + \sin 4\theta + \sin 2\theta = 0,$ then $\theta = $