Question

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

Answer 1

$$\begin{gathered} 2{\sin }^2\theta -1=0 \\ \Rightarrow {\sin }^2\theta =\frac{1}{2} \\ \mathrm{i}.\mathrm{e}\sin \theta =\pm \frac{1}{\sqrt{2}}\mathrm{i}.\mathrm{e}\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6};\mathit{}n\in \mathrm{\mathbb{Z}} \\ \mathrm{ie}\theta =\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4} \end{gathered}$$