Đáp án:
a) \(k \in \left\{ {0;1;....;1009} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sin x + \sqrt 3 \cos x = 2\\
\to \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = 1\\
\to \sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3} = 1\\
\to \sin \left( {x + \dfrac{\pi }{3}} \right) = 1\\
\to x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\to x = \dfrac{\pi }{6} + k2\pi \left( {k \in Z} \right)\\
Do:0 < x < 2020\pi \\
\to 0 < \dfrac{\pi }{6} + k2\pi < 2020\pi \\
\to - \dfrac{\pi }{6} < k2\pi < \dfrac{{12119}}{6}\pi \\
\to - \dfrac{1}{{12}} < k < \dfrac{{12119}}{{12}}\\
\to k \in \left\{ {0;1;....;1009} \right\}\\
b)\cos x + \sin x = 1\\
\to \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\
\to \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\
\to \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
TH1: - 50\pi < x < 50\pi \\
\to - 50\pi < k2\pi < 50\pi \\
\to - 25 < k < 25\\
\to k \in \left\{ { - 24; - 23;...;24} \right\}\\
TH2: - 50\pi < \dfrac{\pi }{2} + k2\pi < 50\pi \\
\to - \dfrac{{101\pi }}{2} < k2\pi < \dfrac{{99\pi }}{2}\\
\to - \dfrac{{101}}{4} < k < \dfrac{{99}}{4}\\
\to k \in \left\{ { - 25; - 24; - 23;...;24} \right\}
\end{array}\)
