How do you solve #sin^4x-1=0#?

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Answer 1 · 2016-12-02T21:08:26

The solutions are #{(3pi)/2+2kpi,pi/2+2kpi}#, #kin ZZ#

Let's factorise the expression

#a^2-b^2=(a+b)(a-b)#

#sin^4-1=0#

#(sin^2+1)(sin^2-1)=0#

#(sin^2+1)(sinx+1)(sinx-1)=0#

So,

#sin^2x+1=0# (no solution)

#sinx+1=0# and #sinx-1=0#

#sinx=-1# and #sinx=1#

#x=(3pi)/2+2kpi# and #x=pi/2+2kpi#, #k in ZZ#