How do you solve $2 {sin}^{2} x + 3 sin x + 1 = 0$ $2sin^2x+3sinx+1=0$ and find all solutions in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2016-09-14T19:27:47

$x = \frac{7 π}{6} or \frac{11 π}{6} or \frac{3 π}{2}$ $x= (7pi)/6 or (11pi)/6 or (3pi)/2$

We may identify this as a quadratic equation in $sin x$ $sinx$ and factorise it as a trinomial to obtain the solutions as

$(2 sin x + 1) (sin x + 1) = 0$ $(2sinx+1)(sinx+1)=0$

$thereforesinx=-1/2 or sinx=-1$

$therefore x= (7pi)/6 or (11pi)/6 or (3pi)/2$

How do you solve 2sin^2x+3sinx+1=02sin2x+3sinx+1=02sin^2x+3sinx+1=0 and find all solutions in the interval [0,2pi)[0,2π)[0,2pi)?