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

The real roots of the equation $cos^7x\, +\, sin^4x\, =\, 1$ in the interval $(-\pi, \pi)$ are

If $\cos 2\theta = (\sqrt 2 + 1)\,\,\left( {\cos \theta - \frac{1}{{\sqrt 2 }}} \right)$, then the value of $\theta $ is

Number of solutions of the equation $2^x + x = 2^{sin \ x} + \sin x$ in $[0,10\pi ]$ is -

Let $S=\left\{\theta \in[-\pi, \pi]-\left\{\pm \frac{\pi}{2}\right\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\right\} \text {. }$ If $T =\sum_{\theta \in S } \cos 2 \theta$, then $T + n ( S )$ is equal

The total number of solution of $sin^4x + cos^4x = sinx\, cosx$ in $[0, 2\pi ]$ is equal to