Question

Find all the solutions within interval [0, 2pi)
2 sin^2 x = sin^2 x

Answer 1

@mathmale

Answer 2

Set it equal to zero. Solve for sin x.

Answer 3

2 sin^2x - sinx = 0
I basically clueless at this stage.

Answer 4

correction: 2 sin^2x - sinx^2 = 0

Answer 5

sin^2x

Answer 6

that's correct
lets say sin x =y
so if we replace sinx with y \[\large\rm 2y^2-y^2=0\]
how would you solve this equation ?

Answer 7

Wouldn't we take the square root of both sides before setting it equal to 0?

Answer 8

no. first we need to set it equal to zero.
now as you can the variables of both terms are the same \[\large\rm \rm 2\color{red}{y^2}-1\color{Red}{y^2}=0\]
combine like terms ? ye?

Answer 9

see*

Answer 10

Oh okay. I had it written down on my paper wrong so i didn't see them as like terms.

Answer 11

so sin^2x = 0

Answer 12

yep

Answer 13

Where does it go from there?

Answer 14

how would you get rid of the square(^2)) ???

Answer 15

Take the square root but there's nothing on the right side of the equation

Answer 16

\(\color{blue}{\text{Originally Posted by}}\) @OnePieceFTW
so sin^2x = 0
\(\color{blue}{\text{End of Quote}}\)
^^^^ there is 0 at right side.

Answer 17

and yes take square root both sides

Answer 18

\[2 \sin ^2x-\sin ^2x=0,~or~\sin ^2x=0,2\sin ^2x=2*0=0,1-\cos 2x=0,
\cos 2x=1\]
\[\cos 2x=1=\cos 0,\cos 2 \pi,\cos 4 \pi,2x=0,2 \pi, 4 \pi,x=0,\pi,2 \pi\]