Question

Could someone explain to me how to evaluate an equation for theta?

Answer 1

\[2\sin ^{2}\theta=1\]

Answer 2

\(\bf 2sin^2(\theta)=1\implies sin^2(\theta)=\cfrac{1}{2}\implies \sqrt{sin^2(\theta)}=\sqrt{\cfrac{1}{2}}\\ \quad \\
sin(\theta)=\cfrac{1}{\sqrt{2}}\\ \quad \\
{\color{blue}{ \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{2} }}\\ \quad \\
sin(\theta)=\cfrac{1}{\sqrt{2}}\implies sin(\theta)=\cfrac{\sqrt{2}}{2}\qquad \textit{now take }sin^{-1}\textit{ to both sides}\)

Answer 3

Thank you so much, I understand how to do it now!

Answer 4

yw