2 cos^2 theta + 2cos theta -1 =0

Question

find the exact solution of the equation on the interval [0, 2pi)

Answer 1

Let x=cosθ, then:

2x²+2x-1=0,

x=(-2±√(4+8))/4=(-1±√3)/2.

cosθ=(-1±√3)/2, θ=cos^-1((-1±√3)/2). We can eliminate (-1-√3)/2, because this value is less than -1 and cosine is always between -1 and 1. θ=cos^-1((√3-1)/2). 2π-cos^-1((√3-1)/2) is also a solution.

(√3-1)/2=cos(π/6)-sin(π/6).

cos(a+b)=cos(a)cos(b)-sin(a)sin(b).

If a=π/4, and b=π/6, cos(π/4+π/6)=cos(5π/12)=

cos(π/4)cos(π/6)-sin(π/4)sin(π/6)=(1/√2)(½√3-½)=cosθ/√2.

Hence cosθ=cos(5π/12)√2, θ=cos^-1(cos(5π/12)√2) or 2π-cos^-1(cos(5π/12)√2) (the exact answer).

An approximation to this solution is θ=1.1961 or 5.0871 radians (68.53° or 291.47°).