Asked by Queen
                The complete solution for the equation cos cube thita- cos thita=0 , 0<or equal to thita<or equal to 2 pi in radian measure is: x1=, x2=..... , x3=...., x4=..., x5=.... fill in only intergral values
            
            
        Answers
                    Answered by
            oobleck
            
    first, it's "theta" not "thita"
and 0 <= θ <= 2π is a lot less messy
so you want to solve
cos^3θ - cosθ = 0
cosθ (cos^2θ - 1) = 0
cosθ = 0 ... θ = π/2, 3π/2
or
cos^2θ - 1 = 0
cosθ = ±1 ... θ = 0, π , 2π
there are no integer solutions
    
and 0 <= θ <= 2π is a lot less messy
so you want to solve
cos^3θ - cosθ = 0
cosθ (cos^2θ - 1) = 0
cosθ = 0 ... θ = π/2, 3π/2
or
cos^2θ - 1 = 0
cosθ = ±1 ... θ = 0, π , 2π
there are no integer solutions
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