Question

the domain is: 0≤x<2π
solve for solutions of sin^2(x)=1/2

Answer 1

\[\sin ^{2}x = \frac{ 1-\cos2\theta }{ 2 }\]

Using that identity, see if you can then solve it more easily : )

Answer 2

Thank you :)

Answer 3

Good luck ^_^

Answer 4

Did you get 3pi/4, 7pi/4, pi/4, and 5pi/4?
@Psymon

Answer 5

Looks good to me :3

Answer 6

:D
Would it be alright if I asked for your help on similar problems if I need it?

Answer 7

go for it, I dont mind :3

Answer 8

2cos^2(x)-1-cos(x)=0
0≤x<2π
@Psymon

Answer 9

This would just be normal factoring. Let cosx = u and solve for u.

\[2u ^{2} - cosu - 1 = 0\]

Answer 10

Oops, should just be u, not cos u, lol.

Answer 11

\[2u ^{2} - u - 1 = 0\]

Answer 12

Well, I simplified it to

\[\cos 2 x - \cos x = 0\]

Answer 13

Yeah, you're overthinking it xD You have to accomplish two things with these. If possible, you must have the same trig function. Everything sines, everything cosines, etc. Secondly, you must have the same angle. You cannot solve for much of anything if one term is 2x and the other is x. So you actually created more work for yourself by changing the problem. All ya had to do was keep the original problem and treat it liek a quadratic by factoring it :P

Answer 14

Oh my gosh, I hate this...
So, I factor it, then what?

Answer 15

Well, trig is pretty awkward for people who are new to it, so it's okay. Well, what did you get after you factored?

Answer 16

(x+1/2) and (x-1)

Answer 17

Right, so now set each of those equal to 0 and solve for x. Just make sure to replace x with cos x once youredone with that.

Answer 18

And, that's it?

Answer 19

Well, you should now have cosx = 1 and cosx = -1/2. Now you just need the values of cos that satisfy those values.

Answer 20

So, 0, 2pi/3, 4pi/3?

Answer 21

Looks good to me. Sorry for late reply, been hopping around.

Answer 22

It's alright, thank you so much for your help. You would think I would get this, figuring that I'm going to become an AP Calc student in 2 weeks... >.<

Answer 23

Youd be amazedhow many calculus students are clueless about trig. I've seen it first hand. I've worked with a lotof calculus students and they'd get to some derivative, tangent line problem, something with trig in it and they'd want to avoid it like the plague. Trig is just a bit abstract from what most are used to. You'll get it, though :3

Answer 24

I just wasn't taught it. I'm kind of learning as I go along. c:

Answer 25

Fair enough. I took the class and I felt like I wasnt taught it either, haha. I kinda had to relearn it really.

Answer 26

How did you do it?

Answer 27

Well, Id seen it already from a horrible teacher, so that still helped some. Then a lot of learning I think came just from taking calc and calc II plus doing math tutoring. The more math you see, the more you understand a lot of things in general. Substitutions in trig started making more sense, I practiced a little more when I felt the need and yeah. And the same with calculus, calc II, tutoring, practice, I just got a better understanding of it not only by studying it, but by just continuing math in general.

Answer 28

And I mean that I tutor math, lol. Its what I do to pay for school really. It forces you to review all the time so you can help others, so it helps out a lot ^_^

Answer 29

I see. Well, thank you so much for your help tonight. :)
Have a good day!

Answer 30

Night and good luck :3