Question

2(2cos4x+1)cosx=1 

Answer 1

2(2cos(4x)+1)cos(x)=1,

2cos(x)(2(2cos²(2x)-1)+1)=1,

2cos(x)(4cos²(2x)-1)=1,

2cos(x)(2cos(2x)-1)(2cos(2x)+1)=1,

2cos(x)(2(2cos²(x)-1)-1)(2(2cos²(x)-1)+1)=1,

2cos(x)(4cos²(x)-3)(4cos²(x)-1)=1.

Let y=2cos(x) then y(y²-3)(y²-1)=1,

y⁵-4y³+3y-1=0=(y²+y-1)(y³-y²-2y+1).

Let f(y)=y⁵-4y³+3y-1, f'(y)=5y⁴-12y²+3 then, using Newton's iterative method:

y_n+1=y_n-f(y_n)/f'(y_n).

Let y₀=0, then y₁=-(-1)/3=⅓, y₂=5/12, ..., y=2cos(x)=0.4450418679, so x=±3π/7+2πn, where n is an integer.

Let y₀=1, then y₁=¾, y₂=73/111, ..., y=2cos(x)=0.6180339888, so x=±2π/5+2πn.

Let y₀=2, then y₁=13/7, ..., y=2cos(x)=1.801937736, so x=±π/7+2πn.

Let y₀=-1, then y₁=-5/4, ..., y=2cos(x)=-1.246979604, so x=±5π/7+2πn.

Let y₀=-2, then y₁=-9/5, ..., y=2cos(x)=-1.618033989, so x=±4π/5+2πn.

The substitutions for y₀ appear to cover all 5 roots of the quintic equation, implying that the 5 generic solutions for x (in radians) represent all the solutions for x. The solutions are all rational fractions of π, so this suggests that another method is likely to exist which gives the solutions without involving solving equations of higher degree than 2. Two of the solutions for y involve the Golden Ratio which are the roots of the quadratic y²+y-1=0 (see above) and the other three solutions for y are roots of the cubic y³-y²-2y+1=0.

From the above factorisation of the quintic we can also write:

y²+y-1=0, that is, 4cos²(x)+2cos(x)-1=0, 2(cos(2x)+1)+2cos(x)-1=0,

2cos(2x)+2cos(x)+1=0, or 2cos(2x)+2cos(x)=-1.

y=(-1±√5)/2, hence cos(x)=(-1±√5)/4,

so x=2π/5+2πn or 2π/5+2πn and 2cos(4π/5)+2cos(2π/5)+1=0, or 2cos(8π/5)+2cos(4π/5)+1=0.

y³-y²-2y+1=0, that is, 8cos³(x)-4cos²(x)-4cos(x)+1=0,

4cos(x)(2cos²(x)-1)-(4cos²(x)-2+1)=0,

4cos(x)cos(2x)-(2cos(2x)+1)=0,

4cos(x)cos(2x)-2cos(2x)-1=0, or 2cos(2x)(2cos(x)-1)=1.

So 2cos(2π/7)(2cos(π/7)-1)=1, 2cos(6π/7)(2cos(3π/7)-1)=1, 2cos(10π/7)(2cos(5π/7)-1)=1.