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

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DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1\[2\sin ^{2}(x)\cos(x)=\cos(x)\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2Set it equal to zero and then take out the common factor

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1\[2\sin ^{2}(x)\cos(x)\cos(x)=0\] like this?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2so what is the common factor ?

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1At 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
 one year ago
Best ResponseYou've already chosen the best response.2right!!! i don't why i was thinking that there is a sign between cos and sin you're right!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\[2\sin^2=0\] now solve for sin

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2hmm i don't think so... cuz if we divide both sides by cos(x)we will get just sin function

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2i would not divide by cos.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\[\huge\rm 2\sin^2(x)\cos(x)=\cos(x)\] is this your question ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2yeah okay let's say sin=x and cos = y \[\huge\rm 2x^2y=y\] you would subtract y both sides not divide

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1yes, but since the question is asking to solve the equation, isn't it asking for when the equations intercept?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2ye they are asking for solutions (x,y) < exact value of cos and sin between 0 to 2pi

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\] \[2\sin^2(x)=1\] 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
 one year ago
Best ResponseYou've already chosen the best response.2well you confused me lol let me confirm

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2btw i'll go with 2nd method 1) subtract 2) find common factor :)

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1okay I understood how to do it from the comment before it was deleted

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2ahh freckles you here i was looking for u but you wer offline ;~;

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[2 \sin^2(x)\cos(x)\cos(x)=0 \\ \cos(x)(2\sin^2(x)1)=0 \\ \cos(x)=0 \text{ or } 2 \sin^2(x)1=0 \\ \cos(x)=0 \text{ or } 2\sin^2(x)=1\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2forgot 1 there facepalm

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1okay, so for cos(x)=0 that would be pi/2 and 3pi/2 correct?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @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 \[\large\rm \color{Red}{2\sin^2\cos(x) \cos(x)=0}\]\[\cos(x)[2\sin^2(x)1]=0\]

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1yes, so the other one would be \[\sin ^{2}(x)=\frac{ 1 }{ 2 }\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2yep but that's sin^2 so take square root both sides

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1\[\sin(x)=\pm \frac{ \sqrt{2} }{ 2 }\] ?

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1okay so that's pi/4, 3pi/4, 5pi/4, and 7pi/4

DanielaJohana
 one year ago
Best ResponseYou've already chosen the best response.1I get it thank you!
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