How do I solve $2sin^2x-1=0$ ?

Question

19208 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-07T21:56:57

$pi/4, (3pi)/4, (5pi)/4, (7pi)/4$ on the range $(0-2pi)$

or to be more exact:

$pi/4 + k*pi/2$ where $k$ is all integers.

$2sin^2x-1=0$

$2sin^2x=1$

$sin^2x=1/2$

$sinx=+-sqrt(1/2)$

$sinx=+-sqrt(2)/2$

Using the unit circle we see:

$sinx=sqrt(2)/2$ at $pi/4 and (3pi)/4$

$sinx=-sqrt(2)/2$ at $(5pi)/4$ and $(7pi)/4$

enter image source here

How do I solve 2sin^2x-1=02sin^2x-1=0?