Solving Trigonomic Equations. How do I go about solving 4sin^4x + 3sin^2x -1 =0?

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Answer 1 · 2018-01-25T12:15:41

#4sin^4x + 3sin^2x -1 =0#

#=>4sin^4x + 4sin^2x-sin^2x -1 =0#

#=>4sin^2x(sin^2x+1) -(sin^2x +1) =0#
#=>(4sin^2x-1)(sin^2x+1) =0#

#(sin^2x+1) =0# not possible

When

#(4sin^2x-1)=0#

#=>2(1-cos2x)=1#

#=>2-2cos2x)=1#

#=>2cos2x=1#

#=>cos2x=1/2=cos(pi/3)#

#=>2x=2npipmpi/3" where "n in ZZ#

#=>x=npipmpi/6" where "n in ZZ#