Question

2cos^2(theta)-sin(theta) =1

Answer 1

from [0,2pi)

Answer 2

using the identity cos^2(theta)=1-sin^2(theta)
2(1-sin^2(theta))-sin(theta)=1
2-2sin^2(theta)-sin(theta)=1
-2sin^2(theta)-sin(theta)+1=0 (factor out -2)
sin^2(theta)+(1/2)sin(theta)-(1/2)=0 (factor)
(sin(theta)+1)(sin(theta)-(1/2))=0
So where sin(theta)=-1 and sin(theta)=(1/2) or 3pi/2 and pi/6. I hope this helps

Answer 3

how did you factor out the -2?

Answer 4

and why wouldn't you subtract -1 instead of add?

Answer 5

and where did the 2 in 2-2sin.... go?

Answer 6

Okay. the first thing I did was replace cos^2(theta) with 1-sin^2(theta). From there I subtracted the 1 from the right to the left making the 2 and 1. Then factored out a negative 2 and multiplied it to the other side (0*-2=0) then factored the quadratic. Does that help any? :/

Answer 7

ok, at the end, why isn't it (sin(theta)-1)(sin(theta)+1/2)=0?

Answer 8

jk

Answer 9

I understand. Thank you!

Answer 10

You're welcome! Just let me know if you have any questions! I wrote than in a hurry sorry :(

Answer 11

just one more, isn't 3pi/2 -1? not 1/2?

Answer 12

It is, I just wrote them in the wrong order.

Answer 13

the sin of it I mean.

Answer 14

oh ok.

Answer 15

So the 1/2 comes from the pi/6 and the -1 comes from the 3pi/2. Sorry to write it backwards.

Answer 16

no worries. thank you!

Answer 17

No problem! :D