So if you want to find the nth term so
1st term=1
2nd term=5
3rd term=25
we notice that 5^1=5 and 5^2=25 , but they are on the second ones so
the expression is
the nth term=5^(n-1)
so 1st term=5^(1-1)=5^0=1
2nd term=5^(2-1)=5^1=5
3rd term=5^(3-1)=5^2=25
and so on
the expression is 5^(n-1)
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
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Answer:
whats the answer?
Step-by-step explanation:
no sabo
