hey. show 2sin(pi/14) is a solution to
x^3-x^2-2x+1=0
and hint: let x=2sinu and sin(7u)=1
i can't seem to solve when i sub in the x value. how can i relate this to the seventh root of inifinity. i multiplu by sin^4 but they it just doesnt work.
To show that 2sin(pi/14) is a solution to the equation x^3 - x^2 - 2x + 1 = 0, let's follow the hint given: let x = 2sin(u) and sin(7u) = 1.
First, substitute x = 2sin(u) into the equation:
(2sin(u))^3 - (2sin(u))^2 - 2(2sin(u)) + 1 = 0
Simplify this expression:
8sin^3(u) - 4sin^2(u) - 4sin(u) + 1 = 0
Now, let's relate this expression to sin(7u) = 1:
We know that sin(7u) can only be 1 when 7u is either π/2 or (7u - π/2) is an integer multiple of 2π.
So, we have two cases to consider:
Case 1: 7u = π/2
In this case, u = π/14. Substitute u = π/14 into the equation we simplified earlier:
8sin^3(π/14) - 4sin^2(π/14) - 4sin(π/14) + 1 = 0
Case 2: (7u - π/2) is an integer multiple of 2π
In other words, 7u - π/2 = 2nπ, where n is an integer. Rearrange the equation to solve for u:
7u = π/2 + 2nπ
u = (π/2 + 2nπ)/7
Substitute this value of u into the equation we simplified earlier:
8sin^3((π/2 + 2nπ)/7) - 4sin^2((π/2 + 2nπ)/7) - 4sin((π/2 + 2nπ)/7) + 1 = 0
These two cases cover all possible values of u that satisfy sin(7u) = 1. Therefore, the equation x^3 - x^2 - 2x + 1 = 0 has solutions when x = 2sin(π/14) and x = 2sin((π/2 + 2nπ)/7), where n is an integer.
Note that relating this to the seventh root of infinity may not be necessary in solving this equation.