TRIGO! Am i doing it right?

[kinuel8091]

The range is from 0≤x<2pi
the equation is
2sin³(x) = sin(x)
2sin³(x)-sin(x)=0

Then factor out sin x
sin(x)(2sin²(x)-1)=0

Then equate both to 0
sin(x)= 0
x=arcsin 0
x=0,pi

2sin²(x)-1=0
sin(x)=√(1/2) or √(2)/2
x=pi/4 and 3pi/4

so, x= 0,pi,pi/4,3pi/4

 

[daon2]

Almost. You missed something when taking the square root

\(\displaystyle \sin^2(x) = \frac{1}{2} \implies |\sin(x)| = \frac{\sqrt{2}}{2}\)

 

[kinuel8091]

ohhhh

Almost. You missed something when taking the square root

\(\displaystyle \sin^2(x) = \frac{1}{2} \implies |\sin(x)| = \frac{\sqrt{2}}{2}\)

I forgot the positive and negative value of √2/2
THANKS!! =))