Question

@Vocaloid

Answer 1

\[2 \sin^2 \theta + 3 \cos \theta - 3 = 0\]

How would you solve for theta here. Just trying to see if I'm doing it the sloppy way.

Answer 2

\[0 \le \theta < 2 \pi\]

Answer 3

i'm not voocoolood but yes using pythagorean identity and solving the quadratic and checking for extraneous solutions is the way to go

Answer 4

Just going through a study guide.

Answer 5

noob

Answer 6

one week

Answer 7

@Shadow have you attempted to use cos^2 + sin^2 = 1 to re-write everything in terms of cos?

Answer 8

Yeah I get 0 as my only answer. Is that what you get?

Answer 9

\[\cos \theta (-2 \cos \theta + 3) - 1 = 0\]

Cause

Answer 10

Has to be within the unit circle and the other one exceeds it.

Answer 11

check your factoring again

(i'm gonna use x instead of theta cause i'm lazy)

2(1-cos^2(x)) + 3cos(x) - 3 = 0

distribute, factor

Answer 12

okay give me a min

Answer 13

any progress so far?

2(1-cos^2(x)) + 3cos(x) - 3 = 0

distributing gives us

2 - 2cos^2(x) + 3cos(x) - 3 = 0

combine the 2 and -3, then factor

Answer 14

2 - 2cos^2(x)+ 3cos(x) - 3
-2cos^2(x) + 3cos(x) - 1
Then I get what I got earlier
cos(x)(-2cos(x) + 3) - 1 = 0

Answer 15

there's a better way to factor it, can you try factoring it like a trinomial?

-2x^2 + 3x - 1?

Answer 16

result should look something like (cos(x)+____)(cos(x)+_____) or some variant

Answer 17

I got it.

(-2cos(x) + 1)(cos(x) - 1)

Answer 18

how is vocaroo not sick of you yet shaddy

Answer 19

good then set each part equal to 0 and solve for cos

-2cos(x) + 1 = 0
cos(x) - 1 = 0

Answer 20

also pikabat is like my bestie how dare u

Answer 21

i'm just emulating my behavior back during OS days lol

Answer 22

shad don't let me derail you did you make progress? or do you need access to my wolfram alpha show steps :^]

Answer 23

I was able to get it. Pi/3 and 0

Answer 24

Just on mobile and too lazy to close.

Answer 25

hm. you sure about that?

I get cos(x) = 1 and cos(x) = 1/2 which should produce three distinct cos values (remember there are two values where cos(x) = 1/2)

Answer 26

5pi/3 ?

Answer 27

good so pi/3, 5pi/3, 0

Answer 28

so uh if you're done i'm gonna close this whoo