Dada a função $f$

Dada a função $f$ tal que $$f(x) = \sqrt 3 \operatorname{sen} x + \cos x$$

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1. Ora,
   $$\begin{array}{*{20}{l}}
   {f(x)}& = &{2 \times \left( {\frac{{\sqrt 3 }}{2}\operatorname{sen} x + \frac{1}{2}\cos x} \right)} \\
   {}& = &{2 \times \left( {\operatorname{sen} x \times \cos \left( {\frac{\pi }{6}} \right) + \cos x \times \operatorname{sen} \left( {\frac{\pi }{6}} \right)} \right)} \\
   {}& = &{2\operatorname{sen} \left( {x + \frac{\pi }{6}} \right)}
   \end{array}$$
   Logo, $a = 2$ e $\alpha  = \frac{\pi }{6}$.
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2. Ora,
   $$\begin{array}{*{20}{l}}
   {f(x) = 1}& \Leftrightarrow &{2\operatorname{sen} \left( {x + \frac{\pi }{6}} \right) = 1} \\
   {}& \Leftrightarrow &{\operatorname{sen} \left( {x + \frac{\pi }{6}} \right) = \frac{1}{2}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x + \frac{\pi }{6} = \frac{\pi }{6} + 2k\pi }& \vee &{x + \frac{\pi }{6} = \pi  – \frac{\pi }{6} + 2k\pi ,k \in \mathbb{Z}}
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 2k\pi }& \vee &{x = \frac{{2\pi }}{3} + 2k\pi ,k \in \mathbb{Z}}
   \end{array}}
   \end{array}$$