by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
Answer:
1. Positive, 1+2=3
2. Negative, -1-2=-3
Step-by-step explanation:
If you look at both in a graphing perspective, the point (1,2) is in Quadrant I. likewise, adding 2 to the x-coordinate will also result in the point (3,2), also in Quadrant I, where the x coordinate is positive. The point (-1,2) is in Quadrant II, and adding -2 to the x coordinate keeps it in Quadrant II, where the x-coordinate is negative.
Answer:
y = -6
x = 1
Step-by-step explanation:
-x + 2y = -13
-x - 2y = 11
Sum the equations:
-x -x = -2x
+2y - 2y = 0
-13 + 11 = -2
then
-2x = -2
x = -2/-2
x = 1
from the first eq.
-x + 2y = -13
-1 + 2y = -13
2y = -13 + 1
2y = -12
y = -12/2
y = -6
check:
from the second eq.
-x -2y = 11
-1 -(2*-6) = 11
-1 -(-12) = 11
-1 + 12 = 11
I think the answer is 0
8*3=24
24 *0=0
I'd say C...
I did the same thing today.
