Cubic equation leads to trigonometric equation to obtain its real roots

Exercise:

Solve in $\mathbb{R}\,,\,\,8{{x}^{3}}-6x+1=0$

Solution: we have $8{{x}^{3}}-6x+1=0\Leftrightarrow 8{{x}^{3}}-6x=-1\Leftrightarrow 6x-8{{x}^{3}}=1\Leftrightarrow 3x-4{{x}^{3}}=\frac{1}{2}$

Let $x=\sin \theta \Leftrightarrow 3\sin \theta -4{{\sin }^{3}}\theta =\frac{1}{2}$

But $3\sin \theta -4{{\sin }^{3}}\theta =3\sin \theta -4{{\sin }^{2}}\theta \sin \theta =\sin \theta \left( 3-4{{\sin }^{2}}\theta  \right)=\sin \theta \left( 1+2-4{{\sin }^{2}}\theta  \right)$

$=\sin \theta \left( 1+2\left( 1-2{{\sin }^{2}}\theta  \right) \right)=\sin \theta \left( 1+2\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2{{\sin }^{2}}\theta  \right) \right)=\sin \theta \left( 1+2\cos 2\theta  \right)$

$=2\sin \theta \cos 2\theta +\sin \theta =2\sin \theta \left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta  \right)+\sin \theta =2\sin \theta {{\cos }^{2}}\theta -2\sin \theta {{\sin }^{2}}\theta +\sin \theta $

$=2\sin \theta \cos \theta \cos \theta -2\sin \theta {{\sin }^{2}}\theta +\sin \theta =\sin 2\theta \cos \theta -\sin \theta \left( 2{{\sin }^{2}}\theta -1 \right)$

$=\sin 2\theta \cos \theta +\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)=\sin 2\theta \cos \theta +\sin \theta \cos 2\theta =\sin \left( 2\theta +\theta  \right)=\sin 3\theta $

Thus $3\sin \theta -4{{\sin }^{3}}\theta =\frac{1}{2}\Leftrightarrow \sin 3\theta =\frac{1}{2}=\sin \left( \frac{\pi }{6} \right)\Leftrightarrow 3\theta =\frac{\pi }{6}+2k\pi \,\,or\,\,3\theta =\pi -\frac{\pi }{6}+2k\pi \,\,\,\,\,,\,\,\,k\in \mathbb{Z}$

$\Rightarrow \theta =\frac{\pi }{18}+\frac{2k\pi }{3}\,\,\,\,\,or\,\,\,\,\,\theta =\frac{\pi }{3}-\frac{\pi }{18}+\frac{2k\pi }{3}\,\,\,\,\,\,,\,\,\,\,k\in \mathbb{Z}$

So $x=\sin \left( \frac{\pi }{18} \right)=0.173648\,\,\,,\,\,x=\sin \left( \frac{5\pi }{18} \right)=0.766044\,\,\,or\,\,\,x=\sin \left( \frac{\pi }{18}-\frac{2\pi }{3} \right)=-0.939693$