Question

Solve each equation over [0,2pi)
4cos^4x-13cos^2x+3=0

Answer 1

`\bf 4cos^4(x)-13cos^2(x)+3=0\impliedby \textit{notice, is really just a quadratic} \\\\\\ 4[~~[cos(x)]^2~~]^2-13[cos(x)]^2+3=0 \\\\\\\ [4cos^2(x)-1][cos^2(x)-3]=0\\\\ -------------------------------\\\\ 4cos^2(x)-1=0\implies 4cos^2(x)=1\implies cos^2(x)=\cfrac{1}{4} \\\\\\ cos(x)=\pm\sqrt{\cfrac{1}{4}}\implies cos(x)=\pm\cfrac{√(1)}{√(4)}\implies cos(x)=\pm\cfrac{1}{√(2)}`


`\bf cos(x)=\pm\cfrac{√(2)}{2}\implies \measuredangle x= \begin{cases} (\pi )/(4)\\\\ (3\pi )/(4)\\\\ (5\pi )/(4)\\\\ (7\pi )/(4) \end{cases}\\\\ -------------------------------\\\\ cos^2(x)-3=0\implies cos^2(x)=3\implies cos(x)=\pm√(3) \\\\\\ cos(x)\approx \pm 1.7`

now, recall that, cosine for any angle has a range from -1 to 1, anything beyond that, is an invalid value, thus certainly 1.7 is so. Meaning, there's no such angle(s) for the second root.