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2sec^2 x + tan^2 x - 3 = 0

 

answers: pi/6, 5pi/6, 7pi/6, 11pi/6

Guest Apr 30, 2017
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\(2\sec^2x+\tan^2x-3=0\\~\\ 2\cdot\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}-\frac{3\cos^2x}{\cos^2x}=0\\~\\ \frac{2+\sin^2x-3\cos^2x}{\cos^2x}=0\\~\\ 2+\sin^2x-3\cos^2x=0 \qquad , \, x\neq\frac{\pi}{2}+n\pi \\~\\ 2 + (1-\cos^2x)-3\cos^2x=0 \\~\\ 3 - 4\cos^2x=0 \\~\\ \cos^2x=\frac34 \\~\\ \cos x=\frac{\sqrt3}{2}\)

 

From looking at a unit circle, the angles within [0, 2π) that have a cosine of \( \frac{\sqrt3}{2} \) are \( \frac{\pi}{6} \) and \( \frac{11\pi}{6} \)   smiley

hectictar  Apr 30, 2017

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