How do you solve $2sin^4x-sin^2x=0$ ?

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Answer 1 · 2016-11-21T06:49:44

The solutions are $S= {npi,(pi/4+(npi)/2)}$ , where $n$ is an integer.

Let's factorise the LHS of the equation

$sin^2x(2sin^2x-1)=0$

$sin^2x=0$ $=>sinx=0$

$x=0, pi, 2pi, npi$

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

$sin^2x=1/2$

$sinx=+-1/sqrt2$

$x=pi/4, 3pi/4, 5pi/4, 7pi/4$

The solutions are $x in {npi,(pi/4+(npi)/2)}$ , where $n$ is an integer.

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