Question

I am having difficulties with this problem-----

Give the most general solutions to the equation: 2sinxcosx - sin(2x)cos(2x) = 0

a. Simplify the first expression using the double-angle identity for sine.

b. Factor the left side of the equation.

c. Solve the factored equation.

Answer 1

\[2\sin x \cos x-\sin2x \cos2x=2\sin x \cos x- 2\sin x \cos x(1-2\sin ^{2} x)=2\sin x \cos x-2\sin x \cos x+4\sin ^{3}x \cos x=4\sin ^{3} x \cos x\]
so, \[4\sin ^{3} x \cos x=0\]
so we get, \[\sin x=0, \cos x=0\]
for \[\sin x=0,\] we get \[x=n \pi \] for \[n \in Z\]
for \[\cos x=0, x= (n+1/2) \pi, n \in Z\]

Answer 2

first equation is not visible fully, it will be:
\[2 \sin x \cos x- 2 \sin x \cos x+ 4 \sin ^{3} x \cos x= 4 \sin ^ {3} x \cos x\]