Tan (4x-π÷3) =√3với sin(2x-40° =√3÷2

Đáp án:

Giải thích các bước giải: \(\begin{array}{l} a)\,\,\tan \left( {4x - \frac{\pi }{3}} \right) = \sqrt 3 \\ \Leftrightarrow \tan \left( {4x - \frac{\pi }{3}} \right) = \tan \frac{\pi }{3}\\ \Leftrightarrow 4x - \frac{\pi }{3} = \frac{\pi }{3} + k\pi \\ \Leftrightarrow 4x = \frac{{2\pi }}{3} + k\pi \\ \Leftrightarrow x = \frac{\pi }{6} + \frac{{k\pi }}{4}\\ b)\,\,\sin \left( {2x - 40^\circ } \right) = \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - 40^\circ } \right) = \sin 60^\circ \\ \Leftrightarrow \left[ \begin{array}{l} 2x - 40^\circ = 60^\circ + k.360^\circ \\ 2x - 40^\circ = 120^\circ + k.360^\circ \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 2x = 100^\circ + k.360^\circ \\ 2x = 160^\circ + k.360^\circ \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 50^\circ + k.360^\circ \\ x = 80^\circ + k.360^\circ \end{array} \right. \end{array}\)

\[\begin{array}{l}
a)tan(4x - \frac{\pi }{3}) = \sqrt 3 \\
 \Leftrightarrow tan(4x - \frac{\pi }{3}) = tan\frac{\pi }{3}\\
 \Leftrightarrow 4x - \frac{\pi }{3} = \frac{\pi }{3} + k\pi \\
 \Leftrightarrow 4x = \frac{\pi }{3} + k\pi \\
 \Leftrightarrow x = \frac{\pi }{6} + \frac{{k\pi }}{4}\\
b)sin(2x - {40^O}) = \sqrt {32} \\
 \Leftrightarrow sin(2x - {40^O}) = sin{60^O}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - {40^O} = {60^O} + k{.360^O}\\
2x - {40^O} = {120^O} + k{.360^O}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = {100^O} + k{.360^O}\\
2x = {160^O} + k{.36^O}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = {50^O} + k{.360^O}\\
x = {80^O} + k{.360^O}
\end{array} \right.\\

\end{array}\]