Giải các phương trình sau: a) sin(2x +15 độ) =√2/2 b) sin(2x-π/5)= -√3/2 28/07/2021 Bởi Rylee Giải các phương trình sau: a) sin(2x +15 độ) =√2/2 b) sin(2x-π/5)= -√3/2
Đáp án: a) $\left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$ b) $\left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$ Giải thích các bước giải: a) $\sin(2x + 15^o) = \dfrac{\sqrt2}{2}$ $\Leftrightarrow \sin(2x + 15^o) = \sin45^o$ $\Leftrightarrow \left[\begin{array}{l}2x +15^o = 45^o + k.360^o\\2x + 15^o = 135^o + k.360^o\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$ b) $\sin\left(2x – \dfrac{\pi}{5}\right) = -\dfrac{\sqrt3}{2}$ $\Leftrightarrow \sin\left(2x – \dfrac{\pi}{5}\right) = \sin\left(-\dfrac{\pi}{3}\right)$ $\Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{5} = -\dfrac{\pi}{3} + k2\pi\\2x – \dfrac{\pi}{5}= \dfrac{4\pi}{3} + k2\pi\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$ Bình luận
Đáp án: Vậy phương trình có họ nghiệm là: ${a)x = {{15}^0} + k{{180}^0}\left( {k \in Z} \right);x = {{60}^0} + k{{180}^0}\left( {k \in Z} \right)}$ ${b)x = – \dfrac{\pi }{{15}} + k\pi \left( {k \in Z} \right);x = \dfrac{{23\pi }}{{30}} + k\pi \left( {k \in Z} \right)}$ Giải thích các bước giải: $\begin{array}{l}a)\sin \left( {2x + {{15}^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}2x + {15^0} = {45^0} + k{360^0}\\2x + {15^0} = {135^0} + k{360^0}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = {15^0} + k{180^0}\\x = {60^0} + k{180^0}\end{array} \right.\\b)\sin \left( {2x – \dfrac{\pi }{5}} \right) = – \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}2x – \dfrac{\pi }{5} = – \dfrac{\pi }{3} + k2\pi \\2x – \dfrac{\pi }{5} = \dfrac{{4\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{{15}} + k\pi \\x = \dfrac{{23\pi }}{{30}} + k\pi \end{array} \right.\end{array}$ Bình luận
Đáp án:
a) $\left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$
b) $\left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
a) $\sin(2x + 15^o) = \dfrac{\sqrt2}{2}$
$\Leftrightarrow \sin(2x + 15^o) = \sin45^o$
$\Leftrightarrow \left[\begin{array}{l}2x +15^o = 45^o + k.360^o\\2x + 15^o = 135^o + k.360^o\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$
b) $\sin\left(2x – \dfrac{\pi}{5}\right) = -\dfrac{\sqrt3}{2}$
$\Leftrightarrow \sin\left(2x – \dfrac{\pi}{5}\right) = \sin\left(-\dfrac{\pi}{3}\right)$
$\Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{5} = -\dfrac{\pi}{3} + k2\pi\\2x – \dfrac{\pi}{5}= \dfrac{4\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Đáp án:
Vậy phương trình có họ nghiệm là:
${a)x = {{15}^0} + k{{180}^0}\left( {k \in Z} \right);x = {{60}^0} + k{{180}^0}\left( {k \in Z} \right)}$
${b)x = – \dfrac{\pi }{{15}} + k\pi \left( {k \in Z} \right);x = \dfrac{{23\pi }}{{30}} + k\pi \left( {k \in Z} \right)}$
Giải thích các bước giải:
$\begin{array}{l}
a)\sin \left( {2x + {{15}^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + {15^0} = {45^0} + k{360^0}\\
2x + {15^0} = {135^0} + k{360^0}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = {15^0} + k{180^0}\\
x = {60^0} + k{180^0}
\end{array} \right.\\
b)\sin \left( {2x – \dfrac{\pi }{5}} \right) = – \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{5} = – \dfrac{\pi }{3} + k2\pi \\
2x – \dfrac{\pi }{5} = \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{{15}} + k\pi \\
x = \dfrac{{23\pi }}{{30}} + k\pi
\end{array} \right.
\end{array}$