solve the equation 2sin^2 (x)cos(x)=cos(x) on the interval [0,2pi)

- DanielaJohana

solve the equation 2sin^2 (x)cos(x)=cos(x) on the interval [0,2pi)

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- DanielaJohana

\[2\sin ^{2}(x)\cos(x)=\cos(x)\]

- Nnesha

Set it equal to zero
and then take out the common factor

- DanielaJohana

\[2\sin ^{2}(x)\cos(x)-\cos(x)=0\]
like this?

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## More answers

- Nnesha

yes right :)

- Nnesha

so what is the common factor ?

- DanielaJohana

At the beginning, couldn't I have just divided both sides by cos(x)? that would have gotten rid of it on both sides and have set it equal to zero
@Nnesha

- Nnesha

right!!!
i don't why i was thinking that there is a sign between cos and sin
you're right!

- Nnesha

\[2\sin^2=0\]
now solve for sin

- Nnesha

wait hmm

- Nnesha

hmm i don't think so...
cuz if we divide both sides by cos(x)we will get just sin function

- Nnesha

i would not divide by cos.

- Nnesha

\[\huge\rm 2\sin^2(x)\cos(x)=\cos(x)\] is this your question ?

- DanielaJohana

@Nnesha yes

- Nnesha

yeah
okay let's say sin=x
and cos = y
\[\huge\rm 2x^2y=y\] you would subtract y both sides not divide

- DanielaJohana

yes, but since the question is asking to solve the equation, isn't it asking for when the equations intercept?

- Nnesha

ye they are asking for solutions (x,y) <-- exact value of cos and sin
between 0 to 2pi

- Nnesha

\[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\]
\[2\sin^2(x)=1\]
but if you subtract cos(x) both sides you will get \[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)]=0\]
so after that you should set both function equal to zero in this way
you will get cos and sin both function value

- Nnesha

well you confused me lol
let me confirm

- Nnesha

btw i'll go with 2nd method
1) subtract
2) find common factor :)

- Nnesha

o.O

- DanielaJohana

okay I understood how to do it from the comment before it was deleted

- freckles

we can't divide both sides by 0
and cos(x) can be 0
--
but yeah you will end up with two equations to solve in the end:
cos(x)=0 and 2 sin^2(x)=1

- Nnesha

ahh freckles you here
i was looking for u
but you wer offline ;~;

- freckles

\[2 \sin^2(x)\cos(x)-\cos(x)=0 \\ \cos(x)(2\sin^2(x)-1)=0 \\ \cos(x)=0 \text{ or } 2 \sin^2(x)-1=0 \\ \cos(x)=0 \text{ or } 2\sin^2(x)=1\]

- Nnesha

forgot 1 there facepalm

- DanielaJohana

okay, so for cos(x)=0 that would be pi/2 and 3pi/2 correct?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
\[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\]
\[2\sin^2(x)=1\]
but if you subtract cos(x) both sides you will get \[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)]=0\]
so after that you should set both function equal to zero in this way
you will get cos and sin both function value
\(\color{blue}{\text{End of Quote}}\)
correction
\[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)-1]=0\]

- Nnesha

ye that's right

- DanielaJohana

yes, so the other one would be \[\sin ^{2}(x)=\frac{ 1 }{ 2 }\]

- Nnesha

yep but that's sin^2 so take square root both sides

- DanielaJohana

\[\sin(x)=\pm \frac{ \sqrt{2} }{ 2 }\] ?

- Nnesha

yes right

- DanielaJohana

okay so that's pi/4, 3pi/4, 5pi/4, and 7pi/4

- Nnesha

nice

- DanielaJohana

I get it thank you!

- Nnesha

np :)

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