Question 739045
Find all solution in the interval [0,2pi) by algebraically: 2sin(^2)*2x+5sin2x-3=0
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{{{2sin^2(2x)+5sin(2x)-3=0}}}
{{{(2sin(2x)-1)(sin(2x)+3)=0}}}
..
(2sin(2x)-1)=0
sin(2x)=1/2
2x=π/6, 5π/6 (in Q1 and Q2 where sin>0)
x=π/12, 5π/12
..
or
(sin(2x)+3)=0
sin(2x)=-3(reject, (-1 ≤ sin(2x) ≤ 1)