Question #9a852

1 Answer
Mar 21, 2017

2sin²x-3sinx + 1 = 0

=>2sin²x-2sinx-sinx + 1 = 0

=>2sinx(sinx-1)-(sinx - 1) = 0

=>(sinx-1)(2sinx - 1) = 0

when

sinx-1=0

=>sinx=1=sin(pi/2)

=>x= npi+(-1)^npi/2" where " n in ZZ

when

2sinx-1=0

=>sinx=1/2=sin(pi/6)

=>x= npi+(-1)^npi/6" where " n in ZZ