Question

3-3sinx-2cos^2x = 0
Find all solutions in the domain [-pi,pi]

Answer 1

Just change cos2x into cos^2x-sin^2x and solve quadratic equation

Answer 2

\[3-3\sin x-2(1-\sin ^2x)=0\]

Answer 3

I just need someone to check answers with I got \[-\pi/2, \pi/2, 2\pi/3, -2\pi/3\]

Answer 4

\[3-3\sin x-2+2\sin ^2x=0\]

Answer 5

Not what I got.

Answer 6

wait I mean

Answer 7

\[-\pi/2 , \pi/2 , 5\pi/6, -5\pi/6 , -\pi/6, \pi/6\]

Answer 8

Not what I got.

Answer 9

What did you get?

Answer 10

Once I factored the quadratic I got (2sinx-1) (sinx-1)=0

Answer 11

\[\frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2}\]

Answer 12

Yes. So sinx=1/2 or 1

Answer 13

The domain is [-pi,pi] though?

Answer 14

Which is the same thing as [0,2pi]

Answer 15

Wait I don't get how that's the same?

Answer 16

The domain is negative pi to pi? How is it 0 to 2pi?

Answer 17

That means give all the solutions between - pi and pi . -pi to 0 would include quadrants 3 and 4. 0 to pi would include quadrants 1 and 2

Answer 18

But -pi/2 and -5pi/6 is between them?

Answer 19

The sin of -pi/2 is -1
The sin of -5pi/6 is -1/2

Answer 20

So those cannot be solutions.

Answer 21

And besides -pi/2 is the same as 3pi/2

Answer 22

And -5pi/5 is the same as 7pi/6

Answer 23

-5pi/6. Sorry for the typo

Answer 24

sin of -1 is 3pi/2 though

Answer 25

Do you mean that the sin of 3pi/2 is -1?

Answer 26

yeah, you said
"The sin of -pi/2 is -1"

Answer 27

How can they both be?

Answer 28

And that is correct because the sin of -pi/2 is the same as the sin of 3pi/2

Answer 29

Let me draw you a picture.