Question

*PLEASE CHECK MY WORK*
Give the most general solutions to the equation below - PLEASE and THANKS?
2sinxcosx - sin(2x)cos(2x) = 0

Part I: Simplify the first expression using the double-angle identity for sine.
Part II: Factor the left side of the equation.
Part III: Solve the factored equation.

Answer 1

Part I
The double angle identity for sine states that sin(2x) = 2sinxcosx
sin(2x) - sin(2x)cos(2x) = 0

Part II
sin(2x)(1 - cos(2x)) = 0

Part III
Either sin(2x) = 0 or
cos(2x) = 1

For sin(2x) = 0
2x = n(pi)
x = n(pi/2)

For cos(2x) = 1
2x = n(pi)
x = n(pi/2)

Answer 2

Is this correct?

Answer 3

Looks good to me!

Answer 4

\[\cos(2x)=1 \rightarrow 2x=0+2n \pi \rightarrow x= n \pi \] you promised to stop rushing @ilfy214 :P

Answer 5

NO! You're right! Uh oh :(

Answer 6

it's okay, just stop rushing and you will be fine ;)

Answer 7

I will.. try! Better?\[\cos(2x)=1\]\[2x=2πn\]\[x=πn\]

Answer 8

yes of course this is better, just try not simplify your steps in such a way that makes these little mistakes happen

Answer 9

Okay! And thank you :D