# The most general value of θ satisfying 2sin

Question:

The most general value of θ satisfying 2sin2θ – 1 = 0 is ______________.

Solution:

$2 \sin ^{2} \theta-1=0$

$\Rightarrow \sin ^{2} \theta=\frac{1}{2}$

i. e $\sin \theta=\pm \frac{1}{\sqrt{2}} \quad$ i. e $\theta=n \pi \pm \frac{\pi}{6} ; n \in \mathbb{Z}$

ie $\theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$