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

1. DanielaJohana

$2\sin ^{2}(x)\cos(x)=\cos(x)$

2. Nnesha

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

3. DanielaJohana

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

4. Nnesha

yes right :)

5. Nnesha

so what is the common factor ?

6. 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

7. Nnesha

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

8. Nnesha

$2\sin^2=0$ now solve for sin

9. Nnesha

wait hmm

10. Nnesha

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

11. Nnesha

i would not divide by cos.

12. Nnesha

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

13. DanielaJohana

@Nnesha yes

14. Nnesha

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

15. DanielaJohana

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

16. Nnesha

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

17. 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

18. Nnesha

well you confused me lol let me confirm

19. Nnesha

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

20. Nnesha

o.O

21. DanielaJohana

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

22. 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

23. Nnesha

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

24. freckles

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

25. Nnesha

forgot 1 there facepalm

26. DanielaJohana

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

27. 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$

28. Nnesha

ye that's right

29. DanielaJohana

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

30. Nnesha

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

31. DanielaJohana

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

32. Nnesha

yes right

33. DanielaJohana

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

34. Nnesha

nice

35. DanielaJohana

I get it thank you!

36. Nnesha

np :)