What is the value of x?

Mathematics

1 answer:

Andreas93 [3]3 years ago

7 0

What is the value of x of what equation?

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

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

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

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

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Answer:

Step-by-step explanation:

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