Question

give the most general solutions to equations

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

A. simplify the first expression using double angle identity for sine.
B. factor the left side of equatioin
C. solve the factored equation

Answer 1

i dont know how to simplify it, factor ir nor solve

Answer 2

2sinxcosx = sin(2x)

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

sin(2x)-sin(2x)cos(2x)=0
now factor out sin(2x)

sin(2x)[1 - cos(2x)] = 0
sin(2x) = 0..since sin(0) = 0.. 2x = 0
x = 0

1- cos(2x) = 0

1 = cos(2x)...since cos(0) =1....2x = 0
x = 0

Answer 3

\[2\sin x \cos x-2\sin x \cos x(1-2\sin ^2x)=0\]
\[2\sin x \cos x[1-(1-2\sin ^2x)]=0\]
\[2\sin x \cos x(2\sin^2x)=0\]
\[2\sin x=0, \cos x=0,,2\sin ^2x=0\]

Answer 4

Solve each equation.

Answer 5

how do you solve for the second equation you have listed?