The number of solutions of the equation cos l | vedclass

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

Question

$\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$

$x \in[-3 \pi, 3 \pi]$

$4\left(\cos ^{2}\left(\f

Vedclass · Accepted Answer

$7$

Vedclass · Answer

$\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$

$x \in[-3 \pi, 3 \pi]$

$4\left(\cos ^{2}\left(\frac{\pi}{3}\right)-\sin ^{2} x\right)=\cos ^{2} 2 x$

$4\left(\frac{1}{4}-\sin ^{2} x\right)=\cos ^{2} 2 x$

$1-4 \sin ^{2} x=\cos ^{2} 2 x$

$1-2(1-\cos 2 x)=\cos ^{2} 2 x$

let $\cos 2 x = t$

$-1+2 \cos 2 x=\cos ^{2} 2 x$

$t ^{2}-2 t +1=0$

$( t -1)^{2}=0$

$t =1 \quad \cos 2 x =1$

$2 x =2 n \pi$

$x = n \pi$

$n =-3,-2,-1,0,1,2,3$

The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is

Similar Questions

If ${\sin ^2}\theta = \frac{1}{4},$ then the most general value of $\theta $ is

The sum of solutions in $x \in (0,2\pi )$ of the equation, $4\cos (x).\cos \left( {\frac{\pi }{3} - x} \right).\cos \left( {\frac{\pi }{3} + x} \right) = 1$ is equal to

$sin 3\theta = 4 sin\, \theta \,sin \,2\theta \,sin \,4\theta$ in $0\, \le \,\theta\, \le \, \pi$ has :

The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for

The set of values of $x$ satisfying the equation,${2^{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 4}}} \right)}}$ $- 2$${\left( {0.25} \right)^{\frac{{{{\sin }^2}\,\left( {x\,\, - \,\,{\textstyle{\pi \over 4}}} \right)}}{{\cos \,\,2x}}}}$ $+ 1 = 0$, is :