SOLUTION: how to solve 2sin^2x-3sinx+1=0

Question 374824: how to solve 2sin^2x-3sinx+1=0
Answer by vasumathi(46) (Show Source): You can put this solution on YOUR website!
solve 2sin^2x-3sinx+1=0
Solution : Let sinx = u
plug in this in the given equation
2u^2-3u+1 = 0
This canbe factored as follows...
(2u-1) ( u-1)= 0
2u-1 = 0 -> 2sinx - 1 = 0-----> sinx = 1/2---> x = pie/6 Since this is periodic with a period 2 pie we have (2*n*pie+pie/6) as a complete solution
now ...
u-1 = 0 --->sinx-1=0-> sinx = 1-------->x = pie/2 .Since this is periodic with a period 2 pie we have (2*n*pie+pie/2) as a complete solution
So the solutions are 1)(2*n*pie+pie/6)
2)(2*n*pie+pie/2)

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