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shusha [124]
3 years ago
5

15 tan^3 x=5 tan x Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If

there is no solution, enter NO SOLUTION.)
Mathematics
1 answer:
ANTONII [103]3 years ago
8 0
\bf 15tan^3(x)=5tan(x)\implies 3tan^3(x)=tan(x)
\\\\\\
3tan^3(x)-tan(x)=0\implies tan(x)[3tan^2(x)-1]=0
\\\\\\
\begin{cases}
tan(x)=0\\\\
\measuredangle x=0,\pi \\
----------\\
3tan^2(x)-1=0\\
tan^2(x)=\frac{1}{3}\\
tan(x)=\pm\sqrt{\frac{1}{3}}\\\\
tan(x)=\pm\frac{1}{\sqrt{3}}\\\\
\measuredangle x=\frac{\pi }{6}, \frac{5\pi }{6},\frac{7\pi }{6}, \frac{11\pi }{6}
\end{cases}
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So with this, we will be using proportions. To solve for x, our proportion will be \frac{x}{6} =\frac{6}{x-9}


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Next, replace -9x with 3x-12x x^2+3x-12x-36=0


Next, factor x^2+3x and -12x-36 separately: x(x+3)-12(x+3)=0 . After, rewrite it as (x-12)(x+3)=0


Next, solve for x separately in the two parentheses:

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Since we can't have a negative length, x = 12.



To solve for y, our proportion will be \frac{y}{9} =\frac{12}{y}


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