2021/050) Solve the equation $4x^6-6x^2 + 2\sqrt{2}=0$

Because it has $x^6$ and $x^2$ term we feel that it could be transformed to a trigonometric equation By  putting $x^2=a\cos\,y$

We get $4a^3\cos^3y - 6a\cos\,y + 2\sqrt{2}=0$

Comparing with $4cos^3z - 3\cos\,z$ the coefficients being proportional we get

$\frac{4a^3}{4}= \frac{6a}{3}$ or $a =\sqrt{2}$

Putting $x^2=\sqrt{2}\cos\,y$

we get $4 * 2 \sqrt{2}\cos^3 y - 6\sqrt{2} \cos\,y + 2\sqrt{2} = 0$

or  $4 \sqrt{2}\cos^3 y - 3 \cos\,y + 1 = 0$

or $\cos\,3y = -1 =\cos\, \pi$

so 3 solutions $3y = \pi$ or $3y= 3\pi$ or $3y = 5\pi$

or $x^2 = \sqrt{2} \cos\,\frac{\pi}{3}$   or  $x^2 = \sqrt{2} \cos\,\pi$ or  $x^2 = \sqrt{2} \cos\,\frac{5\pi}{3}$

giving solution $x = \pm \frac{1}{\sqrt[4]{2}}$(double roots)  or $x= \pm  \sqrt[4]{2}i$

Alternative solution

because there is a $\sqrt{2}$ term and $x^2$ and $x^6$ so we can get rid of $\sqrt{2}$ by putting $x = a \sqrt{2}$ to give

$8a^3\sqrt{2} - 6a\sqrt{2} + 2 \sqrt{2} = 0$

or $8a^3-6a+2= 0$

or $4a^3-3a+1 = 0$

by inspection we get a = -1 is a solution so $a+1$ us a factor

we get $4a^3 - 3a + 1 = 4a^2(a+1) - 4a^2 - 3a + 1 = 4a^2(a+1) - 4a(a+1) + a + 1 = (a+1)(4a^2-4a+1) = (a+1)(2a-1)^2$

giving solution a = -1 or $a= \frac{1}{2}$ (double root)

or $x^2 = - \sqrt{2}$ or  $x^2 = \frac{1}{\sqrt{2}}$ (double root)

 giving solution  $x= \pm  \sqrt[4]{2}i$ or $x = \pm \frac{1}{\sqrt[4]{2}}$(double roots)