Question #7d06f

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Answer 1 · 2016-12-02T00:07:01

#x in {pi/6+2pin, pi/2+2pin, (5pi)/6+2pin}# where #n in ZZ#

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

#=> (sin(x)-1)(2sin(x)-1) = 0#

#=> sin(x)-1 = 0 or 2sin(x)-1 = 0#

#=> sin(x) = 1 or sin(x) = 1/2#

#sin(x) = 1 <=> x = pi/2+2pin, n in ZZ#

#sin(x) = 1/2 <=> x = pi/6+2pin or x = (5pi)/6 + 2pin, n in ZZ#

Taken together, we get our solution set:

#x in {pi/6+2pin, pi/2+2pin, (5pi)/6+2pin}# where #n in ZZ#