Mathematics, 02.08.2019 04:00 miltonC4191
Solve the equation on the interval [0,2π] 16sin^5 x + 2sin x = 12sin³ x the hint says there are ten solutions, show work so i can learn!
Answers: 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}\\\\ -------------------------------](/tpl/images/0160/4056/b9be8.png)
![\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}](/tpl/images/0160/4056/9c6de.png)
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Solve the equation on the interval [0,2π] 16sin^5 x + 2sin x = 12sin³ x the hint says there are ten...
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