What are the general solution of equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x-sinx-1=0$ ?

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Answer 1 · 2016-08-01T12:53:37

Roots are given by $x = n π - {(- 1)}^{n} (\frac{π}{6}}$ $x=npi-(-1)^n(pi/6}$

and $x = 2 n π + \frac{π}{2}$ $x=2npi+pi/2$ , where $n$ $n$ is an integer

Roots of $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x-sinx-1=0$ can be obtained by factorizing the function.

$2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x-sinx-1=0$

$\Leftrightarrow 2 {sin}^{2} x - 2 sin x + sin x - 1 = 0$ $hArr2sin^2x-2sinx+sinx-1=0$

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

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

and roots of $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x-sinx-1=0$ are given by

Hence, either $2 sin x + 1 = 0$ $2sinx+1=0$ i.e. $sin x = - \frac{1}{2}$ $sinx=-1/2$ and

$x = n π - {(- 1)}^{n} (\frac{π}{6}}$ $x=npi-(-1)^n(pi/6}$

or $sin x - 1 = 0$ $sinx-1=0$ i.e. $sin x = 1$ $sinx=1$ and

$x = 2 n π + \frac{π}{2}$ $x=2npi+pi/2$ , where $n$ $n$ is an integer

What are the general solution of equation 2sin^2x-sinx-1=02sin2x−sinx−1=02sin^2x-sinx-1=0?