Question

Help me please SOMEONE!!!
cotx-sin2x=0
How do i find the exact solutions in the interval [0,2pi)

Answer 1

do you know what sin and cos are at that interval?

Answer 2

No, just says solve it algebraically in which the solution must be in the given interval

Answer 3

Ahh! accident

Answer 4

oh okay

Answer 5

i didnt mean to delete that.

Answer 6

sin2x-cotx=0
2sinxcosx-cotx=0 (son2x=2sinxcosx - double angle formula)
2sinxcosx-(cosx/sinx)=0 (cotx=cosx/sinx)
sinx(2sinxcosx-(cosx/sinx))=0(sinx)
2sin^2xcosx-cosx=0 (do not cancel out the two cosx's, because you would be losing solutions that way!)
cosx(2sin^2x-1)=0

cosx=0 and 2sin^2x-1=0
for cosx=0, you get x=pi/2 and 3pi/2 (in that interval)
for 2sin^2x-1=0:
2sin^2x=1
sin^2x=1/2
sinx=plus minus rad (1/2)
**rad (1/2) = (rad2)/2 and you know that sin(pi/4)=(rad2)/2 and variations and sin(5pi/4)= - (rad2)/2 and variations
so x= pi/4, 3pi/4, 5pi/4, 7pi/4 in the interval

Answer 7

Replace cot x by (cos x/sin x) and sin 2x by 2sin x*cos x.
cos x/sin x - 2sin x*cox x = cos x(1/sin x - 2sin x) = cot x (1 - 2sin^2 x) =
= cot x*cos 2x = 0.
Solve 2 basic trig equations:
a) cot x = 0 --> x = Pi/2,
b) cos 2x = 0 --> 2x = Pi/2 and 2x = 3Pi/2
x = Pi/4 x = 3Pi/4