Posted by Anonymous on Wednesday, February 22, 2012 at 10:25pm.
Solve over the indicated interval:
2sin^2x=sqrt2sinx , [180,180)
don't how to get x by itself
please help and thank you

Trig Help (URGENT)  Reiny, Wednesday, February 22, 2012 at 10:57pm
let sinx = y, so you get
2y^2 = √(2y)
square both sides
4y^4 = 2y
4y^4  2y = 0
2y(2y^3  1) = 0
y = 0 or y^3 = 1/2  y = (.5)^(1/3) = appr .7937
if y = 0, then sinx = 0
x = 180°, 0° , 180°
if y = .7937 , then sin x = + .7937 (x must be in I or II)
x = 52.53 or 18052.53 = 127.47°
but we squared, so our answers MUST be verfied
if x = 180 or 180
LS = 0 , RS = 0 , ✔
if x = 0, ✔
if x = 52.53°
LS = 1.2599 , RS = 1.2599 ✔
if x = 127.47
Ls = 1.2599 , RS = 1.2599 ✔
All 5 answers above are correct
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