First slide

Trigonometric equations

Question

If $x \in (0, \frac{π}{2})$ , one solution of $\frac{\sqrt{3} - 1}{\sin x} + \frac{\sqrt{3} + 1}{\cos x} = 4 \sqrt{2}$ is $\frac{π}{12}$ , $t h e o t h e r s o l u t i o n i s$ $\frac{λ π}{36}$ where $λ =$

Difficult

A

11
B

12
C

13
D

14

Solution

$\begin{array}{l} G i v e n e q u a t i o n i s (\sqrt{3} + 1) \sin x + (\sqrt{3} - 1) \cos x = 4 \sqrt{2} \sin x \cos x \\ \Rightarrow \frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x + \frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x = 2 \sin x \cos x \\ \Rightarrow \sin x \cos \frac{π}{12} + \cos x \sin \frac{π}{12} = \sin 2 x \\ \Rightarrow \sin (x + \frac{π}{12}) = \sin 2 x \\ \Rightarrow 2 x = n π + {(- 1)}^{n} (x + \frac{π}{12}) \\ \Rightarrow x = \frac{π}{12} o r \frac{11 π}{36} \end{array}$

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