Question

The equation $x^3-\frac{3}{4}x=-\frac{\sqrt{3}}{8}$ is satisfied by

• A. $x=\cos \left(\frac{5\pi }{18}\right)$
• B. $x=\cos \left(\frac{7\pi }{18}\right)$
• C. $x=\cos \left(\frac{23\pi }{18}\right)$
• D. $x=\cos \left(\frac{17\pi }{18}\right)$

Answer 1

Answer: (A), (B), (D)

$x=\cos \left(\frac{5\pi }{18}\right)$
$x=\cos \left(\frac{7\pi }{18}\right)$
$x=\cos \left(\frac{17\pi }{18}\right)$

$x^3-\frac{3}{4}x=-\frac{\sqrt{3}}{8}$
Let $x\Rightarrow \cos \theta$
${\cos }^3\theta -\frac{3}{4}\cos \theta =\frac{-\sqrt{3}}{8}$
$\Rightarrow 4{\cos }^3\theta -3\cos \theta =\frac{-\sqrt{3}}{2}$
$\cos 3\theta =\frac{-\sqrt{3}}{2}\Rightarrow 3\theta =2n\pi +\frac{5\pi }{6}$ or $2n\pi +\frac{7\pi }{6}$
$\Rightarrow \theta =\frac{1}{18}(12\pi n\pm 5\pi )$
$\therefore \theta =\frac{5\pi }{18},\frac{7\pi }{18},\frac{17\pi }{18}$