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:
E. This polynomial could be factored by using grouping or the perfect squares methods.
Step-by-step explanation:
x^2 + 2x + 1
There is no greatest common factor
This is a perfect square
a^2 + 2ab+ b^2 = ( x+1)^2
We can factor this by grouping
x^2 + 2x + 1
(x^2 +x) + (x+1)
x( x+1) + x+1
Factor out x+1
( x+1) ( x+1)
This is not the difference of squares since there is no subtraction
Answer:
10, because 10-3 is 7, and 7 is greater than 6.
Answer:
0?
Step-by-step explanation:
