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 180-52.53 = 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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