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8 sec2 x + 4 tan2 x − 12 = 0

 Jul 17, 2014

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 #2
avatar+99352 
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$$\begin{array}{rlll}
8sec^2x+4tan^2x-12&=&0\\\\
\frac{8}{cos^2x}+4(\frac{1}{cos^2x}-1)-12&=&0\\\\
\frac{8}{cos^2x}+\frac{4}{cos^2x}-4-12&=&0\\\\
\frac{12}{cos^2x}-16&=&0\\\\
\frac{12}{cos^2x}&=&16\\\\
\frac{3}{cos^2x}&=&4\\\\
\frac{3}{4}&=&cos^2x\\\\
cos^2x&=&\frac{3}{4}\\\\
cos\;x&=&\frac{\pm \sqrt3}{2}\\\\
x&=&n\pi\pm \frac{\pi}{6}\qquad \mbox{where}\;\;n \in Z\\\\

\end{array}$$

.
 Jul 18, 2014
 #1
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(2 sin2 x − 1)(3 tan2 x − 1) = 0

 Jul 17, 2014
 #2
avatar+99352 
+10
Best Answer

$$\begin{array}{rlll}
8sec^2x+4tan^2x-12&=&0\\\\
\frac{8}{cos^2x}+4(\frac{1}{cos^2x}-1)-12&=&0\\\\
\frac{8}{cos^2x}+\frac{4}{cos^2x}-4-12&=&0\\\\
\frac{12}{cos^2x}-16&=&0\\\\
\frac{12}{cos^2x}&=&16\\\\
\frac{3}{cos^2x}&=&4\\\\
\frac{3}{4}&=&cos^2x\\\\
cos^2x&=&\frac{3}{4}\\\\
cos\;x&=&\frac{\pm \sqrt3}{2}\\\\
x&=&n\pi\pm \frac{\pi}{6}\qquad \mbox{where}\;\;n \in Z\\\\

\end{array}$$

Melody Jul 18, 2014

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