Question

Solve the equation on the interval [0,2π]

Answer 1

$\bf 16sin^5(x)+2sin(x)=12sin^3(x) \\\\\\ 16sin^5(x)+2sin(x)-12sin^3(x)=0 \\\\\\ \stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\ -------------------------------\\\\ 2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x= \begin{cases} 0\\ \pi \\ 2\pi \end{cases}\\\\ -------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0 \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \begin{array}{lcccl} & 8 sin^4& -6 sin^2(x)& +1\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16} \\\\\\ sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)= \begin{cases} \frac{1}{2}\\\\ \frac{1}{4} \end{cases}$

$\bf \measuredangle x= \begin{cases} sin^{-1}\left( \frac{1}{2} \right) sin^{-1}\left( \frac{1}{4} \right) \end{cases}\implies \measuredangle x= \begin{cases} \frac{\pi }{6}~,~\frac{5\pi }{6}\\ ----------\\ \approx~0.252680~radians\\ \qquad or\\ \approx~14.47751~de grees\\ ----------\\ \pi -0.252680\\ \approx 2.88891~radians\\ \qquad or\\ 180-14.47751\\ \approx 165.52249~de grees \end{cases}$