Finding Values of $x$ satisfying $\max{(3 \cos \alpha + 4 \cos \Bigl( \alpha - x \Bigl)} = 7$

Question

The question in the question set says to find values of $x$ that satisfy the equation: $$\max _ {\forall \;\alpha \;\epsilon\; R} \Biggl( 3 \cos \alpha + 4 \cos \Bigl( \alpha - x \Bigl) \Biggl) = 7$$ I exoanded the inner bracket of $$4\cos \Bigl(\alpha-x \Bigl)$$ To $$ 4\cos \alpha \cos x + 4\sin \alpha \sin x$$ Then modifying the expression we get,$$\Bigl( 3 + 4 \cos x \Bigl) \cdot \cos \alpha + 4 \sin x \sin \alpha = 7$$ After this I'm not sure however what to do. I know what the $\max$ function does but I'm not sure how to proceed from here.
Any help would be much appreciated.

Answer 1

Since $$\max\cos x=1,$$ we can have $$\alpha-x=\cos^{-1}1=2k\pi$$ for some integer $k$ so take $$x=\alpha-2k\pi.$$

Answer 2

The expression will take max value $7$ iff $\cos \alpha =1$ and $\cos (\alpha-x)=1$

So $\alpha = 2\pi\cdot k$ and $\alpha -x = 2\pi\cdot l$ where $k,l\in \mathbb{Z}$.

So $ x= 2\pi \cdot m$ where $m$ is arbitrary integer ($m=l-k$).