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

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Answer 1 · 2016-07-09T07:07:35

$x = \frac{3 π}{2}$ $x=(3pi)/2$

By using quadratic formula, you have

$sin x = \frac{1 \pm \sqrt{1 + 24}}{4}$ $sinx=(1+-sqrt(1+24))/4$

$sin x = \frac{1 \pm 5}{4}$ $sinx=(1+-5)/4$

$sin x = - 1 or sin x = \frac{3}{2}$ $sinx=-1 or sinx=3/2$ (this solution is not possible because $sin x \leq 1$ $sinx<=1$ )

then sinx=-1 when $x = \frac{3 π}{2}$ $x=(3pi)/2$

How do you solve 2sin^2x-sinx-3=02sin2x−sinx−3=02sin^2x-sinx-3=0 between the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?