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
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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:
Step-by-step explanation:
x is at (0, 4)
Ok y is your y value from any point on the graph m is your slope x is the x value of any point from the graph and b is your y intercept that is where the line cross in the y line that is the vertical line in the graph
Answer:
(-6,9,-3)
Step-by-step explanation:
-3x -y +z=6
-3x-y+3z =0
x-3z =3
Multiply the second equation by -1
-1 *(-3x-y+3z) =0*-1
3x +y -3z =0
Add this to the first equation
-3x -y +z=6
3x +y -3z =0
----------------------
0 + 0 + -2z = 6
Divide by -2
-2z/-2 = 6/-2
z = 6/-2
z=-3
Take the third equation to find x
x-3z=3
x-3(-3) = 3
x+9=3
Subtract 9 from each side
x+9-9 =3-9
x=-6
Now we need to find y
3x +y -3z =0
3(-6) +y -3(-3) =0
-18 +y +9=0
-9+y =0
Add 9 to each side
-9+9+y = 0+9
y=9
(-6,9,-3)
