How do you solve for 0 ≤ x < 2π using the equation 4sin ² x - 1 = 0 ?

Feb 28, 2018

$\frac{\pi}{6} , \frac{5 \pi}{6} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$

Explanation:

$4 {\sin}^{2} x - 1 = 0 \Rightarrow {\sin}^{2} x = \frac{1}{4} \Rightarrow \sin x = \textcolor{red}{\pm \frac{1}{2}}$
Given,$0 \le x < 2 \pi$

• $0 \le x < \left(\frac{\pi}{2}\right) \Rightarrow \sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right) \Rightarrow x = \textcolor{red}{\frac{\pi}{6}}$
• $\left(\frac{\pi}{2}\right) \le x < \left(\pi\right) \Rightarrow \sin x = \frac{1}{2} = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \left(\frac{5 \pi}{6}\right) \Rightarrow x = \textcolor{red}{\frac{5 \pi}{6}}$
• $\pi \le x < \frac{3 \pi}{2} \Rightarrow \sin x = - \frac{1}{2} = \sin \left(\pi + \frac{\pi}{6}\right) = \sin \left(\frac{7 \pi}{6}\right) \Rightarrow x = \textcolor{red}{\frac{7 \pi}{6}}$
• $\frac{3 \pi}{2} \le x < 2 \pi \Rightarrow \sin x = - \frac{1}{2} = \sin \left(2 \pi - \frac{\pi}{6}\right) = \sin \left(\frac{11 \pi}{6}\right) \Rightarrow x = \textcolor{red}{\frac{11 \pi}{6}}$
Note:The general solution is
$\sin x = \pm \frac{1}{2} \Rightarrow \sin x = \sin \left(\pm \frac{\pi}{6}\right)$
So,
$x = k \pi + {\left(- 1\right)}^{k} \left(\pm \frac{\pi}{6}\right) , k \in Z$
Taking $k = 0 , \pm 1 , \pm 2 , \ldots$we can obtain above answer.