Đáp án + Giải thích các bước giải:
a) $sin(x-60^o)=\dfrac{1}{2}$
⇔ $sin(x-60^o)=sin.30^o$
⇔ \(\left[ \begin{array}{l}x-60^o=30^o+k.360^o\\x-60^o=180^o-30^o+k.360^o\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=90^o+k.360^o\\x=210^o+k.360^o\end{array} \right.\) $(k∈\mathbb{Z})$
c) $cos(x-2)=\dfrac{2}{5}$
⇔ \(\left[ \begin{array}{l}x-2=arccos(\dfrac{2}{5})+k2\pi\\x-2=-arccos(\dfrac{2}{5})+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2+arccos(\dfrac{2}{5})+k2\pi\\x=2-arccos(\dfrac{2}{5})+k2\pi\end{array} \right.\) $(k∈\mathbb{Z})$
d) $cos(2x+\dfrac{\pi}{3})=-\dfrac{1}{2}$
⇔ $cos(2x+\dfrac{\pi}{3})=cos\dfrac{2\pi}{3}$
⇔ \(\left[ \begin{array}{l}2x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\2x+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{array} \right.\) $(k∈\mathbb{Z})$
e) $cos(2x+50^o)=\dfrac{1}{2}$
⇔ $cos(2x+50^o)=cos60^o$
⇔ \(\left[ \begin{array}{l}2x+50^o=60^o+k360^o\\2x+50^o=-60^o+k360^o\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5^o+k180^o\\x=-55^o+k180^o\end{array} \right.\) $(k∈\mathbb{Z})$
