$\bf 15tan^3(x)=5tan(x)\implies 3tan^3(x)=tan(x)
\\\\\\
3tan^3(x)-tan(x)=0\implies tan(x)[3tan^2(x)-1]=0
\\\\\\
\begin{cases}
tan(x)=0\\\\
\measuredangle x=0,\pi \\
----------\\
3tan^2(x)-1=0\\
tan^2(x)=\frac{1}{3}\\
tan(x)=\pm\sqrt{\frac{1}{3}}\\\\
tan(x)=\pm\frac{1}{\sqrt{3}}\\\\
\measuredangle x=\frac{\pi }{6}, \frac{5\pi }{6},\frac{7\pi }{6}, \frac{11\pi }{6}
\end{cases}$

8 0

3 years ago

What is the sum of the first eight terms of the series?

Ray Of Light [21]

Observe that as the series progresses, the term decreases by 1/4. To show this, observe the first four terms of the series below.

-200 = (1/4)(-800)
-50 = (1/4)(-200)
-12.5 = (1/4)(-50)

Since we have a common ratio, r, of 1/4, we can use the properties of a geometric series to find the 8th term of the series.

Recall that to find the sum of the nth term of a geometric series, we have

$S_{n} = a(\frac{1-r^{n}}{1-r})$

where a is the first term of the series and r is the ratio.

So, for the first eight terms, we have

$S_{8} = -800(\frac{1-(\frac{1}{4})^8}{1- \frac{1}{4}})$
$S_{8} \approx -1066.65$

Therefore, the sum of the 8th series is approximately -1066.65.

Answer: -1066.65

3 0

3 years ago

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