Number of solution (s) of the equation { | vedclass

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

Question

$\left( {{{\cos }^2}2x - \cos x{{\cos }^2}5x + {{\cos }^4}\frac{{5x}}{4}} \right) + \frac{{{{\cos }^2}5x\left( {1 - {{\cos }^2}5x} \right)}}{4} = 0$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

$\left( {{{\cos }^2}2x - \cos x{{\cos }^2}5x + {{\cos }^4}\frac{{5x}}{4}} \right) + \frac{{{{\cos }^2}5x\left( {1 - {{\cos }^2}5x} \right)}}{4} = 0$

$\cos 2 x=\frac{\cos ^{2} 5 x}{2} \& \sin 10 x=0$

$x=\frac{n \pi}{10}$

$x=0, \frac{\pi}{10}, \frac{2 \pi}{10^{\prime}} \frac{3 \pi}{10}$

No solution

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

Similar Questions

Find the principal and general solutions of the question $\tan x=\sqrt{3}$.

If $\tan m\theta = \tan n\theta $, then the general value of $\theta $ will be in

Solve $\cos x=\frac{1}{2}$

Values of $\theta (0 < \theta < {360^o})$ satisfying ${\rm{cosec}}\theta + 2 = 0$ are

$cos (\alpha \,-\,\beta ) = 1$ and $cos (\alpha +\beta ) = 1/e$ , where $\alpha , \beta \in [-\pi , \pi ]$ . Number of pairs of $(\alpha ,\beta )$ which satisfy both the equations is