Question

PLEASE HELP. solve over domain -2pi 13 years ago

Answer 1

4sinxcosx=2sin^2x
2sinxcosx=sin^2x

Answer 2

sin2x=sin^2x

Answer 3

then what

Answer 4

help please

Answer 5

answer key has 6 answers

Answer 6

Put it as quadratic equation
\[4sinxcosx=2-2\cos^2x \\ \\ 2\cos^2x+4sinxcosx-2=0 \]
quadratic formula
\[cosx=\frac{-4\pm \sqrt{16-4(2)(-2)}}{2(2)} \\ \\ cosx=\frac{-4 \pm 4 \sqrt{2}}{4} \\ \\ cosx=-1+\sqrt{2}~|~cosx=-1-\sqrt{2}\]
Find "x"

Answer 7

can you solve sin2x=sin^2x or was what I was doing totally off?

Answer 8

sin2x=sin^2x
Using trig identity, sin2x=2sinxcosx, we get
2sinxcosx=sin^2x

Answer 9

yeah i tried to get it all in sin.. how do u factor sin2x=sin^2x that more?

Answer 10

\[2sinxcosx=\sin^2x \\ \\ 0=\sin^2x-2sinxcosx \\ \\ sinx(sinx-2cosx)=0 \\ \\ sinx=0~~~|~~~sinx-2cosx=0 \\ \\ tanx=2\]
Then solve for sinx=0 and tanx=2

Answer 11

thank youu!!!!!