Question

Please help!

Answer 1

@amistre64 @tcarroll010

Answer 2

@Luigi0210

Answer 3

Change the 2sin^2x

Answer 4

How?

Answer 5

Sin^2x+cos^2x=1
cos^2x=1-sin^2x
sin^2x=1-cos^2x

Answer 6

How does that change sin^2x?

Answer 7

Then just distribute the 2

Answer 8

@Loser66 take this?

Answer 9

You guys it's not that difficult

Answer 10

Once I find a trick to these kind of problems It's easy for me but I'm having trouble with this...

Answer 11

sorry @Luigi0210 I got it, but since you start, finish, please

Answer 12

I don't care who helps I just need to get this done. lol

Answer 13

hahaha, wait for him, friend. It must be done under his hand. I believe it

Answer 14

\[\cos2 \theta = 2\sin^2 \theta \]
\[\cos2 \theta= 2(\cos^2 \theta-1)\]
\[0=2\cos^2 \theta-\cos \theta-2\]
\[(2\cos \theta+1)(\cos \theta -1)\]

Answer 15

darn I messed up

Answer 16

yep

Answer 17

66 finish it I have to go back to work

Answer 18

\[cos 2\theta=1-2sin^2\theta\]
\[1-2sin^2\theta = 2sin^2\theta\]
\[1 = 4sin^2\theta\]
\[sin^2\theta= \frac{1}{4}\]
square root both sides
\[sin\theta = \sqrt{\frac{1}{4}}\]
can you step up from now?

Answer 19

sin theta =+/- 1/2, and solve part by part.OK?

Answer 20

hey, you guys don't know what I mean?