Question

The least difference between the roots of the equation $4cosx(2-3si{n}^{2}x)+(cos2x+1)=0(0\le x\le \pi /2)$ is

• A. $\pi /6$
• B. $\pi /4$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

The correct option is A $\pi /6$

$4cos\left(x\right)[2-3si{n}^{2}x]+(cos2x+1)=0$
$4cos\left(x\right)[3co{s}^2-1]+2co{s}^{2}x=0$
Or
$2cos\left(x\right)[6co{s}^{2}x+cosx-2]=0$
Or
$2cos\left(x\right)=0$ and $6co{s}^{2}x+cos\left(x\right)-2=0$.
Or
$x=\frac{(2n+1)\pi }{2}$ and
$cos\left(x\right)=\frac{-1\pm \sqrt{1+4\left(12\right)}}{12}$
$=\frac{-1\pm 7}{12}$
Hence
$cos\left(x\right)=\frac{1}{2}$ and $cos\left(x\right)=\frac{-2}{3}$.
Since $xϵ[0,\frac{\pi }{2}]$ , hence $cos\left(x\right)$ has to be positive.
Thus $x=\frac{\pi }{2},\frac{\pi }{3}$.
Thus
$\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$.