How do you find the solution to $2 {sin}^{2} θ - 1 = 0$ $2sin^2theta-1=0$ if $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2017-01-31T07:27:53

The solutions are $S={45º, 135º, 225º,315º}$

Let's rewrite the equation

$2sin^2theta-1=0$

$2sin^2theta=1$

$sintheta=+-1/sqrt2=+-sqrt2/2$

When $sintheta=sqrt2/2$ , $=>$ , $theta=45º, 135º$

and when $sintheta=-sqrt2/2$ , $=>$ , $theta=225º, 315º$

How do you find the solution to 2sin^2theta-1=02sin2θ−1=02sin^2theta-1=0 if 0<=theta<3600≤θ<3600<=theta<360?