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

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Answer 1 · 2016-08-01T12:53:37

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

and $x=2npi+pi/2$ , where $n$ is an integer

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

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

$hArr2sin^2x-2sinx+sinx-1=0$

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

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

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

Hence, either $2sinx+1=0$ i.e. $sinx=-1/2$ and

$x=npi-(-1)^n(pi/6}$

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

$x=2npi+pi/2$ , where $n$ is an integer

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