Question

1-cos x=sin x
in the range of [0,2pi)

Answer 1

cosx+sinx=1
then
(cosx+sinx)^2=1

Answer 2

can you tell what is the next ?

Answer 3

sin x = ( 1- cos ^2 x) ^(1/2)
( 1- cos ^2 x) ^(1/2) = 1+cos x

Answer 4

(cosx+sinx)^2=1=cos^2(x)+sin^2(x)+2sinx cosx=1
and it is a rule
cos^2?(theta)+sin^2(theta)=1
so
(cos^2x+sin^2x+2sinxcosx=1+sin2x=1
then
sin2x=0

Answer 5

this pic shows the graph of sin2x
u see sin2x=0 in the period (0,2pi) at
x=90,180,270,360 degrees

Answer 6

and x=0 also