Let $\alpha$ be a real number such that $0\leq\alpha\leq\pi$.If $f(x)=\cos x+\cos(x+\alpha)+\cos(x+2\alpha)$ takes some constant number $c$ for any $x\in R$,then find the value of $\lfloor c+\alpha \rfloor$.
I simplified $f(x)=\cos x+\cos(x+\alpha)+\cos(x+2\alpha)=\cos(x+\alpha)(2\cos\alpha+1)$
If $f(x)$ takes the value $c$,then $f(x)=\cos(x+\alpha)(2\cos\alpha+1)=c$
I do not know how to solve it further.
$\text{cos}(x+\alpha)$ is obviously non-constant. $2\text{cos}(\alpha)+1$ is obviously constant. The product of a constant and a non-constant function is always non-constant, except if this constant is zero (because...)
Since this equation is true for all $x$ we can substitute values and see what we get. Substitute $\pi$ and $0$.
Equation 1: $c = 1 + cos(\alpha) + cos(2\alpha)$
Equation 2: $c = -1 - cos(\alpha) - cos(2\alpha)$
Using these two we get $c=0$
$cos(\alpha) = {{-1}\over 2}$
$\Rightarrow \alpha = {{2\pi}\over 3}$
Hints:
Look at the expression for $x\mapsto f(x)$ you have obtained. Under what circumstances can it be constant? Your findings will then determine $c$ and $\alpha$.
