Question

One of the roots of the equation $8x^3-6x+1=0$ is

• A. $\cos {10}^o$
• B. $\cos {30}^o$
• C. $\sin {30}^o$
• D. $\cos {80}^o$

Answer 1

The correct option is D $\cos {80}^o$

Given $8x^3-6x+1=0$

$\Rightarrow 2(4x^3-3x)+1=0$

Substitute $x=\cos \theta$

$2\left[4\right(\cos \theta {)}^3-3\left(\cos \theta \right)]+1=0$

$2\cos 3\theta =-1$

$\cos 3\theta =\frac{-1}{2}$

$=\cos {120}^o$ or $\cos {240}^o$ or $\cos {480}^o$

$3\theta =120$ or $240$ or $480$

i.e; $\theta =40$ or $80$ or $160$

$\therefore x=\cos {40}^o$ or $\cos {80}^o$ or $\cos {160}^o$

So, $d)$ option is correct

i.e; $\cos {80}^o$