Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

yes right :)

Answer 5

so what is the common factor ?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

wait hmm

Answer 10

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

Answer 11

i would not divide by cos.

Answer 12

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

Answer 13

@Nnesha yes

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

\[\huge\rm \frac{ 2\sin^2(x)\cancel{\cos(x) }}{ \cancel{\cos(x}) } = \frac{ \cancel{\cos(x) }}{ \cancel{\cos(x)}}\]
\[2\sin^2(x)=1\]
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

Answer 18

well you confused me lol
let me confirm

Answer 19

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

Answer 20

o.O

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

forgot 1 there facepalm

Answer 26

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

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @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
\[\large\rm \color{Red}{2\sin^2\cos(x)- \cos(x)=0}\]\[\cos(x)[2\sin^2(x)-1]=0\]

Answer 28

ye that's right

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

yes right

Answer 33

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

Answer 34

nice

Answer 35

I get it thank you!

Answer 36

np :)