## Babybee24 2 years ago cos^2(x)+cos(x)sin(x)=0 solve for x

1. myininaya

Try factoring

2. Babybee24

I factored a cos(x) out but then im stuck

3. myininaya

cos(x)( ? + ? ) =0 So what should be in the places of those ?'s once you factor out cos(x)?

4. Babybee24

cos(x)(cosx+sinx)

5. myininaya

Yes. Set both factors equal to 0. cos(x)=0 ; cos(x)+sin(x)=0

6. myininaya

Solve cos(x)=0 for x And find the value x so that cos(x) and sin(x) are opposites (i.e cos(x)=-sin(x))

7. Babybee24

cosx=pie/2 ?

8. myininaya

cos(x)=0 for x=pi/2 cos(x) does not equal pi/2 But one value that satisfies cos(x)=0 is x=pi/2 Are there any other values?

9. Babybee24

x=3pi/2

10. myininaya

Ok and you can also say we have x=pi/2+2npi where n is an integer or x=3pi/2+2n pi where n is an integer

11. myininaya

We also need to find the values of x where cos(x) and sin(x) have opposite values.

12. myininaya

|dw:1357920385873:dw| As you can see they will be opposite in quadrant 2 and 4.

13. myininaya

|dw:1357920430907:dw| These points is where cos(x) and sin(x) are opposite. Now what angle are these points at from the initial ray?

14. Babybee24

135 and 315?

15. myininaya

Correct.

16. myininaya

17. Babybee24

18. myininaya

So 135*pi/180 or 315*pi/180

19. myininaya

Don't forget we can go around the circle infinitely many times from those angles.

20. myininaya

So 135pi/180+2npi , 315pi/180+2npi

21. myininaya

Don't forget the other solutions we wrote above. Good job. :)

22. Babybee24

thanks!