Question

If $x = 2k\pi + \frac{\pi}{3} = \frac{\pi}{3}(6k + 1)$ then $k \in Z$

• A. $2\cos\theta = 0$
• B. $\theta = 2n\pi \pm \frac{\pi}{2}$
• C. $\tan(3x - 2x) = \tan x = 1$
• D. $x = n\pi + \frac{\pi}{4}$

Answer 1

Answer: (B)

$\theta = 2n\pi \pm \frac{\pi}{2}$

Answer 2

$\Rightarrow$

$\theta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4} \Rightarrow \theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$ $\tan 3x = 1 \Rightarrow \tan 3x = \tan\frac{\pi}{4}$ ⇒ $\Rightarrow 3x = n\pi + \frac{\pi}{4}$

⇒ $\Rightarrow$.