Question

Solve cos X tan X - sin ^2 X=0 for 0 to 360 degrees...???
A. 0,pi/4,pi
B. 0,pi/2
C. 0,pi/2,pi
D. 0,pi, 3pi/2

I think the answer is D

Answer 1

ok

Answer 2

\[\cos(x)\frac{ \sin(x) }{ \cos(x) } - \sin^2(x) = 0\]

Answer 3

so
\[\sin(x) - \sin^2(x) = 0\]

Answer 4

pull out a sin(x)
\[\sin(x)* [ 1-\sin(x)] = 0\]

Answer 5

so you have 2 equations

sin(x) = 0
and
sin(x) = 1

Answer 6

Recall, the sin value on the unit circle is the y coordinate of any point on the unit circle.

Answer 7

When on the circle, is the Y value 0 or 1?

Answer 8

When the angle is 0 degrees, the point on the unit circle is (1,0)

Answer 9

so the sin(x) = 0, when the angle is 0 degrees

Answer 10

Likewise, if the angle is 180 degrees, you are at the point (-1,0), the y value here is also zero, so
sin(180) also is zero,

Answer 11

Now for sin(x) = 1

The Y value is 1, when you are at 90 degrees, you are at point (0,1) on the unit circle

Answer 12

how is it from 0 to 360 if your answers have \(\pi\) in them ? shouldn't it be \(0\) to \(2\pi\)?

Answer 13

yes

Answer 14

ok, 0 degrees is 0 radians
90 degrees is pi/2 radians
180 degrees is pi radians

Answer 15

ok thank you

Answer 16

and I need help with one more question

Answer 17

For Reference...

Answer 18

so the right answer is c

Answer 19

yes, because at 3/2 pi or 270 degrees, the y value of the point is -1 on the y axis

y = sin(x) = -1
which is not +1

Answer 20

Sin(x) has to be either 0, or +1
Which occurs when the y value on that circle is 0 or +1
Which occurs at 0, pi/2, and pi angles

Answer 21

ok I got it and you explained me very well

Answer 22

cool

Answer 23

Find the ordered triple that represents the vector from A(-3,5,6) to B(-6,8,6). Then find the magnitude of AB?
A. (3,-3,0); 4.24
B. (-9,13,12); 19.85
C. (-3,3,0); 4.24
D. ( -9,3,0); 9.49

Answer 24

this one is pretty easy...

Answer 25

really how can you solve this problem

Answer 26

AB = <(Bx - Ax) , (By - Ay), (Bz - Az)>

Answer 27

The vector is that

Answer 28

The magnitude of vector AB = <x , y, z> is
\[\left| AB \right| = \sqrt{x^2 + y^2 +z^2} = \sqrt{(Bx - Ax)^2 + (By - Ay)^2 + (Bz - Az)^2}\]

Answer 29

so AB = < (-6-(-3)) , (8-5) , (6-6)>

Answer 30

AB = <-3, 3, 0>

\[\left| AB \right| = \sqrt{(-3)^2 + 3^2 + 0^2}\]

Answer 31

So the right answer is c

Answer 32

um