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

The solution of equation ${\cos ^2}\theta + \sin \theta + 1 = 0$ lies in the interval

If $A + B + C = \pi$ & $sin\, \left( {A\,\, + \,\,\frac{C}{2}} \right) = k \,sin,\frac{C}{2}$ then $tan\, \frac{A}{2} \,tan \, \frac{B}{2}=$

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

If $\alpha ,\,\beta ,\,\gamma $ and $\delta $ are the solutions of the equation $\tan \left( {\theta + \frac{\pi }{4}} \right) = 3\,\tan \,3\theta $ , no two of which have equal tangents, then the value of $tan\, \alpha + tan\, \beta + tan\, \gamma + tan\, \delta $ is

The general solution of $a\cos x + b\sin x = c,$ where $a,\,\,b,\,\,c$ are constants