First slide

Trigonometric equations

Question

The number of values of $x \in [0, 4 π]$ satisfying $|\sqrt{3} \cos x - \sin x| \geq 2$ is

Moderate

A

2
B

0
C

4
D

8

Solution

$W e h a v e x \in [0, 4 π] a n d |\sqrt{3} \cos x - \sin x| \geq 2$
$\begin{array}{l} B u t |\sqrt{3} \cos x - \sin x| \leq \sqrt{3 + 1} = 2 \\ T h u s w e m u s t h a v e |\sqrt{3} \cos x - \sin x| = 2 \\ \Rightarrow |\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x| = 1 \\ \Rightarrow |\cos (x + \frac{π}{6})| = 1 \\ \Rightarrow \cos (x + \frac{π}{6}) = 1, - 1 \\ ⇒ x + \frac{π}{6} = 0, 2 π, 4 π, 6 π ..... π, 3 π, 5 π, .... \\ ∴ x = \frac{11 π}{6}, \frac{23 π}{6}, \frac{5 π}{6}, \frac{17 π}{6} (∵ x \in [0, 4 π]) \end{array}$

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