Question

Solve over the indicated interval:
2sin^2x=sqrt2sinx , [-180,180)

don't how to get x by itself

please help and thank you

Answer 1

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