Question

How do you get pi/2 for cosx=0 and pi/6 for (1-2sinx)=0?

(See photo)

Answer 1

Do you know about the unit circle? That is the usual reference tool for sine and cosine equations like this.

Answer 2

cos x = 0 is already in a good form for that. The x-value on the unit circle is for cos t, where t is the angle.

1 - 2 sin x = 0 has to be manipulated a bit to isolate the sin x term. Then we can use the unit circle with that. The y-value on the unit circle is sin t, where t is the angle again.

Answer 3

Yes, I know about the unit circle. I'm in Calculus 2. I'm just unaware on how the value of x resulted from cos(x) = 0 . If I use "cos^-1" to try and solve for the angle, I get zero.

Answer 4

cos (x) = 0
cos^-1 (cos (x)) = cos^-1 (0)
x = cos^-1 (0) You get 0 for cos^-1 (0) ?

Answer 5

I'm just re-arranging the equation to solve for the x-vales, but I do not see how cos(x) = o results in pi/2 . In order to get pi/2, I would need to get a result of 90 degrees for the value of cos(x) = 0 . I don't see any way to get pi/2.

Answer 6

Oh never mind. I re-arranged it wrongly.

Answer 7

I got pi/2

Answer 8

Let me try it for the other equation.

Answer 9

Hmmm I'm getting -pi/4

Answer 10

I did: \[\frac{ \sin ^{-1}(-1) }{ 2 }\]

Answer 11

This confuses me a lot

Answer 12

From 1 - 2 sin (x) = 0
You solved for sin (x) and then took the inverse sine of both sides?

Answer 13

Yes. I think my rearrangement is wrong

Answer 14

for cos(x) = 0 , all I did was" x=cos(^1)(0)

Answer 15

That was understandable

Answer 16

If you need, you can write out your full process here and we can see what the issue is.

Answer 17

I typed it above

Answer 18

Well I guess I could type it again.

Answer 19

I mean, in a step-by-step process. That way, I can tell at what point an issue arose.

Answer 20

\[\frac{ \sin^{-1}(-1)}{ 2 }\] = \[-\frac{ \pi }{ 4 }\]

Answer 21

* starting by solving this equation: 1 - 2 sin (x) = 0, for sin x

Answer 22

That's what I can't do.

Answer 23

It would be a few algebra steps. We first subtract 1 from both sides:
1 - 2 sin (x) = 0
-2 sin (x) = -1
Then we divide both sides by -2
sin(x) = -1/-2
sin(x) = 1/2
This is the kind of step-by-step process I would write out for this problem.
If you understand that part, then we take the inverse sine of both sides to get x by itself.

Answer 24

-___- Yes! That's it!!! Thank You!

Answer 25

I really should practice on this.

Answer 26

I was confused on whether to put the -1 in the parenthesis.

Answer 27

Glad to help! Which step were you putting the -1 in parentheses? Like if you had this situation taking the inverse sine of both sides:
sin(x) = -1/2
You would put absolutely everything inside the parentheses in the inverse sine function:
x = sin^-1 (-1/2)
The equality only holds if you take the inverse of everything on both sides; otherwise, you'll change the value by doing something like -sin^-1(1)/2.

Answer 28

But anyways, definitely practice practice practice! It might sound silly to ask to practice Algebra stuff when you're in Calc 2, but really Algebra is fundamental in any level of Calc! I've helped some people on this site doing more complicated differential equations stuff and their issue really boiled down to a problem in the Algebra. :)

Answer 29

Ahhh I see! (Your point about x=sin(x)^-1(-1/2))

Oh you bet! It isn't the Calculus that people tend to mess-up on--it's the Algebra! It's just so frustrating because it really halts your progress when working on a Calculus problem. Oh well.... Practice.... :P

Thanks again, @AccessDenied !

Answer 30

No problem! Good luck on Calc 2 and happy maths! :)