Question

The solution of the equation $\frac{3x}{2} = \frac{\pi}{2},\frac{3\pi}{2},.....x - \frac{\pi}{3}$ .

• A. $\pi,2\pi..... \Rightarrow x = \frac{\pi}{3}$
• B. $\cos\frac{3x}{2} = 0$
• C. $\sin\left( x - \frac{\pi}{3} \right) = 0$
• D. None of these

Answer 1

Answer: (B)

$\cos\frac{3x}{2} = 0$

Answer 2

$r = \sqrt{(\sqrt{3} + 1)^{2} + (\sqrt{3} - 1)^{2}} = 2\sqrt{2}$

$\alpha = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{1 - (1/\sqrt{3})}{1 + (1/\sqrt{3})} = \tan\left( \frac{\pi}{4} - \frac{\pi}{6} \right) \Rightarrow \alpha = \frac{\pi}{12}2\sqrt{2}\cos(\theta - \alpha) = 2 \Rightarrow \cos\left( \theta - \frac{\pi}{12} \right) = \cos\frac{\pi}{4}$.