Question

solve cosx=sin2x on [0,2pi]

Answer 1

cos(x)=sin(2x)

cos(x)=2*sin(x)*cos(x)

cos(x)-2*sin(x)*cos(x) = 0

cos(x)( 1-2*sin(x) ) = 0

I'll let you finish

Answer 2

I got 1/2=sin(x) so x=\[\pi/6, 5\pi/6\]

Answer 3

that's part of the solution set

Answer 4

Oh! Okay, thanks a bunch!

Answer 5

what's the rest of the solution set

Answer 6

\[11\pi/6, 7\pi/6\] ? is that right or am I not getting this at all?

Answer 7

no idk how you got that

Answer 8

if

cos(x)( 1-2*sin(x) ) = 0

then

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

Answer 9

you already solved 1 - 2sin(x) = 0, so just solve the first equation to get your two other solutions

Answer 10

alright, so it would be \[\pi/2, 3\pi/2\]

Answer 11

yep, those are added to the other two solutions you got

Answer 12

so your four solutions are

pi/6, 5pi/6, pi/2, 3pi/2

Answer 13

Thanks a bunch for the help.

Answer 14

you stop at these 4 because you're restricted to the interval [0, 2pi)

Answer 15

yw