Solve the equation `sin(x/2) + cos(x) - 1 = 0.`
I suppose that "sinx/2+cosx-1=0" means "sin(x/2)+cos(x)-1=0" (it can also mean "(sinx)/2+cos(x)-1=0" and "sin(x/2)+cos(x-1)=0").
To solve this equation, recall the double angle formula `cos(2a) = 1 - 2sin^2(a),` and apply it to `cos(x):` `cos(x) = 1 - 2sin^2(x/2).` This way our equation becomes
`sin(x/2) + (1 - 2sin^2(x/2)) - 1 = 0,` or `sin(x/2) - 2sin^2(x/2) = 0,` or `sin(x/2)(1 - 2sin(x/2)) = 0.`
The product is zero means at least one of factors is zero, i.e. `sin(x/2) = 0` or `sin(x/2) = 1/2.` These equations are well-known and their solutions are
`x/2 = k pi, or x = 2k pi,`
`x/2 = pi/6 + 2k pi, or x = pi/3 + 4k pi,`
`x/2 = (5pi)/6 + 2k pi, or x = (5pi)/3 + 4k pi,`
where `k` is any integer.
At `[0, 4pi],` which is a period of `sin(x/2)+cos(x)-1,` the solutions are `0,` `pi/3,` `(5pi)/3,` `2pi` and `4pi.`
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