Question

Solve the equation cos2x = 1-sinx for x in the interval [0,2pi).
I used the double angle formula to get
cos^2x - sin^2x= 1- sin x
then i moved it all to the left side and set it equal to 0:
cos^2x - sin^2x + sinx - 1= 0.
Have I done this correctly so far? Also where do I go next?

Answer 1

ok from there you can use the identity cos^2x = 1 - sin^2x

Answer 2

so the equation will now be 1 - sin^2x - sin^2x + sinx - 1

Answer 3

then do i combine like terms?

Answer 4

yes you do and you should get a quadratic equation

Answer 5

do i get -2sin^2 + sin =0

Answer 6

yes and now you factorise

Answer 7

then do i get sinx(-2sinx+1)=0

Answer 8

yup

Answer 9

now sinx = 0 and -2 sinx +1=0
and you can solve from there

Answer 10

yupp i did that and do i get 0, pi, pi/6 and 5pi/6

Answer 11

i think you should just get 0 and pi/3 because -2 sinx +1 =0 then -2sinx=-1
then sinx= -1/-2 which will be sinx=1/2 which is 30degrees pi/3

Answer 12

oh you are right for some reason i divided it by 2 instead of -2 thank you!!!!

Answer 13

pi/6 sorry

Answer 14

wait but its a multiple choice question and the answers are:
0, pi
pi/6, 5pi/6
0, pi, 7pi/6, 11pi/6
0, pi, pi/6, 5pi/6

Answer 15

wait i think im right with the answers. because you get sin=0 which is 0, pi and you get sin=1/2 which is pi/6, 5pi/6

Answer 16

yes you are

Answer 17

okay thank you for all your help!

Answer 18

no problem