DanielaJohana
  • DanielaJohana
solve the equation 2sin^2 (x)cos(x)=cos(x) on the interval [0,2pi)
Mathematics
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DanielaJohana
  • DanielaJohana
solve the equation 2sin^2 (x)cos(x)=cos(x) on the interval [0,2pi)
Mathematics
chestercat
  • chestercat
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DanielaJohana
  • DanielaJohana
\[2\sin ^{2}(x)\cos(x)=\cos(x)\]
Nnesha
  • Nnesha
Set it equal to zero and then take out the common factor
DanielaJohana
  • DanielaJohana
\[2\sin ^{2}(x)\cos(x)-\cos(x)=0\] like this?

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Nnesha
  • Nnesha
yes right :)
Nnesha
  • Nnesha
so what is the common factor ?
DanielaJohana
  • 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
  • Nnesha
right!!! i don't why i was thinking that there is a sign between cos and sin you're right!
Nnesha
  • Nnesha
\[2\sin^2=0\] now solve for sin
Nnesha
  • Nnesha
wait hmm
Nnesha
  • Nnesha
hmm i don't think so... cuz if we divide both sides by cos(x)we will get just sin function
Nnesha
  • Nnesha
i would not divide by cos.
Nnesha
  • Nnesha
\[\huge\rm 2\sin^2(x)\cos(x)=\cos(x)\] is this your question ?
DanielaJohana
  • DanielaJohana
Nnesha
  • 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
  • DanielaJohana
yes, but since the question is asking to solve the equation, isn't it asking for when the equations intercept?
Nnesha
  • Nnesha
ye they are asking for solutions (x,y) <-- exact value of cos and sin between 0 to 2pi
Nnesha
  • 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
  • Nnesha
well you confused me lol let me confirm
Nnesha
  • Nnesha
btw i'll go with 2nd method 1) subtract 2) find common factor :)
Nnesha
  • Nnesha
o.O
DanielaJohana
  • DanielaJohana
okay I understood how to do it from the comment before it was deleted
freckles
  • 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
  • Nnesha
ahh freckles you here i was looking for u but you wer offline ;~;
freckles
  • 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
  • Nnesha
forgot 1 there facepalm
DanielaJohana
  • DanielaJohana
okay, so for cos(x)=0 that would be pi/2 and 3pi/2 correct?
Nnesha
  • 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
  • Nnesha
ye that's right
DanielaJohana
  • DanielaJohana
yes, so the other one would be \[\sin ^{2}(x)=\frac{ 1 }{ 2 }\]
Nnesha
  • Nnesha
yep but that's sin^2 so take square root both sides
DanielaJohana
  • DanielaJohana
\[\sin(x)=\pm \frac{ \sqrt{2} }{ 2 }\] ?
Nnesha
  • Nnesha
yes right
DanielaJohana
  • DanielaJohana
okay so that's pi/4, 3pi/4, 5pi/4, and 7pi/4
Nnesha
  • Nnesha
nice
DanielaJohana
  • DanielaJohana
I get it thank you!
Nnesha
  • Nnesha
np :)

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