SOLUTION: What are the possible values of x domain: [0,2pi).
4sin^3(x) + 2sin^2(x) = 2sin(x) + 1
So far I've:
set equal to zero:
4sin^3(x) + 2sin^2(x) + 2sin(x) + 1 = 0
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-> SOLUTION: What are the possible values of x domain: [0,2pi).
4sin^3(x) + 2sin^2(x) = 2sin(x) + 1
So far I've:
set equal to zero:
4sin^3(x) + 2sin^2(x) + 2sin(x) + 1 = 0
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Question 381159: What are the possible values of x domain: [0,2pi).
4sin^3(x) + 2sin^2(x) = 2sin(x) + 1
So far I've:
set equal to zero:
4sin^3(x) + 2sin^2(x) + 2sin(x) + 1 = 0 Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! First of all when you set everything equal to zero you should obtain
Let a = sin x. The equation then becomes
This is equivalent to . We can factor to obtain
Solving, we obtain a = -1/2, sqrt(2)/2, and -sqrt(2)/2. All of these values are between -1 and 1, so we can find sin^-1 of each of these angles to find all values for x:
or
