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.