How do you solve $2sin^2x-sinx-3=0$ between the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-07-09T07:07:35

$x=(3pi)/2$

By using quadratic formula, you have

$sinx=(1+-sqrt(1+24))/4$

$sinx=(1+-5)/4$

$sinx=-1 or sinx=3/2$ (this solution is not possible because $sinx<=1$ )

then sinx=-1 when $x=(3pi)/2$

How do you solve 2sin^2x-sinx-3=02sin^2x-sinx-3=0 between the interval 0<=x<=2pi0<=x<=2pi?