First slide

Trigonometric equations

Question

$\cos^{2} x - 2 \cos x = 4 \sin x - \sin 2 x$ If:

Moderate

A

$\tan (\frac{x}{2}) = - \frac{1}{2}$
B

$\tan x = - \frac{1}{2}$
C

$\tan (\frac{x}{2}) = 2 + \sqrt{5}$
D

$\tan (\frac{x}{2}) = 2 - \sqrt{5}$

Solution

$\begin{array}{l} \cos^{2} x - 2 \cos x = 4 \sin x - 2 \sin x \cos x \\ \Rightarrow (\cos x - 2) (\cos x + 2 \sin x) = 0 \\ \Rightarrow \cos x + 2 \sin x = 0 (∵ \cos x - 2 \neq 0) \\ \Rightarrow \tan x = - \frac{1}{2} \\ \Rightarrow - \frac{1}{2} = \frac{2 \tan (\frac{x}{2})}{1 - \tan^{2} (\frac{x}{2})} \\ \Rightarrow \tan^{2} (\frac{x}{2}) - 4 \tan (\frac{x}{2}) - 1 = 0 \\ \Rightarrow \tan (\frac{x}{2}) = 2 \pm \sqrt{5} \end{array}$

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