How do you solve #sin^4x+2sin^2x-3=0# for #0<=x<=2pi#?

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Answer 1 · 2016-08-16T21:30:34

#x = pi/2, (3pi)/2#

This can be factored as #(x^2 + a)(x^2 + b)#.

#=>(sin^2x + 3)(sin^2x - 1) = 0#

#=>sin^2x = -3 and sin^2x = 1#

#=> sinx = sqrt(-3) and sinx = sqrt(1)#

#=> sinx = O/ and sinx = +-1#

#=> x = pi/2, (3pi)/2#

Hopefully this helps!