Question

The least difference between the roots, in the first quadrant $(0\le x\le \frac{\pi }{2})$ , of the equation
$4\cos x(2-3{\sin }^{2}x)+(\cos 2x+1)=0$, is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{6}$

given equation is
$4\cos x(2-3{\sin }^{2}x)+2{\cos }^{2}x=0$
$2\cos x(-2+6{\cos }^{2}x+\cos x)=0$
$\cos x(2\cos x-1)(3\cos x+2)=0$
$\cos x=0$ or $\cos x=\frac{1}{2}$ or $\cos x=-\frac{2}{3}$(impossible in first quadrant)

so $\theta =\frac{\pi }{2}or\frac{\pi }{3}$

Difference $=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$