Question

Help with solving trig equations

Answer 1

\[2-2\sin^2(x)=\sqrt3\cos(x)\]

Answer 2

so we could start by factoring out 2, 2(1-sin^2)=2cos^2

Answer 3

Since there is no domain, you are looking for general solutions, yes?

Answer 4

ooops forgot to ad dthe interval

Answer 5

the interval is [0,360degrees]

Answer 6

If so, I would start by subtracting cos^2

Then, realize that sin^2=1-cos^2

2-2(1-cos^2(x)-cos(x)sqrt3

I would make cos=x, to make this easier to read

2-2(1-x^2)-x(sqrt3)

Same thing as 2x^2-sqrt3(x)=0

factor

x(2u-sqrt3(x)=0

x=0

x=sqrt3/2

now just get the degrees and solutions based on your interval

Answer 7

"x(2u-sqrt3(x)=0
that's a x instead of a u, my mistake.

Answer 8

phew the x's got me confused for a moment I thought that cos just disappeared lol

Answer 9

you don't have to substitute cos=x, but it can be easier to read sometimes (at least for me)

Answer 10

so our first solution would be 60 degrees?

Answer 11

check again, if cos=sqrt3/2

60 degrees would be 1/2

Answer 12

oh you are right I was looking at the wrong column oops. sqrt3/2=1/2=60 degrees

Answer 13

*30 degrees

Answer 14

wow I failed twice in a row

Answer 15

http://www.pinkmonkey.com/studyguides/subjects/trig/chap2/a16.gif

Answer 16

i shoulda just looked up S:

Answer 17

lol, and the 2nd solution is when cos=0, which is what degrees?

Answer 18

90

Answer 19

yes

Answer 20

How can you tell how many solutions an equation has?

Answer 21

well actually, since you have a domain here...and cos is positive, you can make two more solutions

Answer 22

for the 4th quadrant

Answer 23

360-each of your solutions

Answer 24

so 360-90 and 360-30

Answer 25

yes

Answer 26

so our other two solutions will be 270 and 330

Answer 27

yes

Answer 28

so all together 30,90,270,330 :p

Answer 29

yep!, put it in radians if you like

Answer 30

I think it must be in degrees because the interval is in degrees

Answer 31

up to you :P

Answer 32

Thank you for the help:3

Answer 33

np