The given equation is:
\[ 2 \tan^2 x \cos x - 6 \cos x - 4 \tan^2 x + 12 = 0 \]
Recall that \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\). Substitute this into the equation:
\[ 2 \left(\frac{\sin^2 x}{\cos^2 x}\right) \cos x - 6 \cos x - 4 \left(\frac{\sin^2 x}{\cos^2
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