Earlier we learned that `cos(x/2)=+-sqrt((1+cos x)/2)`, so we have:

`+-sqrt((1+cos x)/2)= 1 + cos x`

Squaring both sides gives:

`(1+cos x)/2=(1+cos x)^2`

`(1+cos x)/2=1+2\ cos x+cos^2x`

`1+cos x=2+4\ cos x+2\ cos^2x`

So we have:

`2\ cos^2 \ x + 3\ cos x + 1 = 0`

`(2\ co\s x + 1)(cos x + 1) = 0`

Solving, we get

`cos x = − 0.5` or `cos x = − 1`

Now `cos x=-1/2` gives `x=(2pi)/3,(4pi)/3`.

However, on checking in the original equation, we note that

`"LHS"=cos((4pi)/3xx1/2)` `=cos((2pi)/3)` `=-1/2`


`"RHS"=1+cos (4pi)/3=+1/2`

So the only solution for this part is `x=(2pi)/3.`

Also, `cos x=-1` gives `x = pi`.

So the solutions for the equation are `x=(2pi)/3or pi.`

A check of the graph of `y=cos x/2-1-cos x` confirms these results:

Graph of trig equation solution

(2π/3 ≈ 2.0944 and π ≈ 3.14).

Please support IntMath!