Question

Solve this trigonometric equation.
2sin^2-1=0

Answer 1

Note that this is the same as \[\cos (2x)=0\]

Answer 2

\[\sin(x)=\pm\frac{\sqrt{2}}{2}\] is a start

Answer 3

how didyou get cos(2x)

Answer 4

Therefore \[x=\pi/4+n*\pi/2, n \in \mathbb{Z}\]

Answer 5

try 2sin^2x = 1 and then sin^2x = 1/2 then go from there

Answer 6

so\[\sin(x) = \sqrt{(1/2)} or \pm1/\sqrt{?}\]

Answer 7

? = 2

Answer 8

but the restriction id 0 to pi/2