Question

Find the solutions of the equations in the interval [0,2pi). cos(x/2)-sinx=0

Answer 1

write it like this: cos(x/2)-sin(x/2+x/2)=0

Answer 2

I can change sinx into sin(x/2+x/2) ?

Answer 3

Yes. then you will have: cos(x/2)-sin(x/2)cos(x/2)-cos(x/2)sin(x/2)=0

Answer 4

now take cos(x/2) out:
cos(x/2) (1-sin(x/2)-sin(x/2) )= 0

Answer 5

or:
cos(x/2) (1-2sin(x/2) )=0

Answer 6

Ah what formula is that because the given formulas i have are that of the Pythagorean theorem, half-angle theorem, power-reducing formulas, and Double-angle formulas

Answer 7

double angle formula

Answer 8

I wrote it like the sum of two equal angles, but the result is still same

Answer 9

this last expretion will be 0 when either cos(x/2) =0 or 2sin(x/2) = 1 or better say
sin(x/2) =1/2

Answer 10

Oh isn't when sin(x/2+x/2) = sinx/2cosx/2+sinx/2cosx/2? not sinx/2cosx/2-sinx/2cosx/2

Answer 11

cos x= 0 at Pi/2 and 3Pi/2, so cos(x/2)=0 at Pi and 3Pi

Answer 12

don't forget the minus sign infront of it -+=-

Answer 13

Oh okay thank you

Answer 14

so now you have to find where sin(x/2)=1/2. Can you do it?

Answer 15

yes, but to make sure is cos(x/2) = 0 just the same as cos(x) = 0

Answer 16

no. cos x= 0 at Pi/2 and 3Pi/2, so cos(x/2)=0 at Pi and 3Pi

Answer 17

Oh woops i accidently multiplied the bottom part by two and not the top part

Answer 18

Thank you i solved it :3

Answer 19

good job

Answer 20

its 5pi/6 X3

Answer 21

which are inside [0,2Pi) ?

Answer 22

since its sin it should be x/2 = pi/6 and 5pi/6 and cos x/2 = pi/2 3pi/2 making my final outcome at x= pi/3, 5pi/3, and pi because 3pi is out of the interval

Answer 23

er wait wasn't that sin because cosx/2=0

Answer 24

and sinx/2=+1/2 at the points on my unit circles says that they are at pi/6 and 5pi/6 o_o

Answer 25

sry . You right. sin5Pi/6 = 1/2

Answer 26

my bad

Answer 27

Yay i'm right XD but thank you for helping me figure this thing out :3

Answer 28

yw