giải PT a, sin(15độ -2x)=-1 b, sin2x+sin5x=0 c, cosx – 3 độ= $\frac{√2}{2}$ d, sin ²x +sin ²3x =1 20/09/2021 Bởi Hailey giải PT a, sin(15độ -2x)=-1 b, sin2x+sin5x=0 c, cosx – 3 độ= $\frac{√2}{2}$ d, sin ²x +sin ²3x =1
Đáp án: $\begin{array}{l}a)\sin \left( {{{15}^0} – 2x} \right) = – 1\\ \Rightarrow {15^0} – 2x = – {90^0} + k{.360^0}\\ \Rightarrow 2x = {105^0} – k{.360^0}\\ \Rightarrow x = 52,{5^0} + k{.180^0}\\b)\sin 2x + \sin 5x = 0\\ \Rightarrow sin2x = – sin5x\\ \Rightarrow sin2x = \sin \left( { – 5x} \right)\\ \Rightarrow \left[ \begin{array}{l}2x = – 5x + k2\pi \\2x = \pi + 5x + k2\pi \end{array} \right.\left( {k \in Z} \right)\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{{k2\pi }}{7}\\x = \dfrac{{ – \pi }}{3} – \dfrac{{k2\pi }}{3}\end{array} \right.\left( {k \in Z} \right)\\c)\cos \left( {x – {3^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\ \Rightarrow \cos \left( {x – {3^0}} \right) = \cos {45^0}\\ \Rightarrow \left[ \begin{array}{l}x – {3^0} = {45^0} + k{.360^0}\\x – {3^0} = – {45^0} + k{.360^0}\end{array} \right.\left( {k \in Z} \right)\\ \Rightarrow \left[ \begin{array}{l}x = {48^0} + k{.360^0}\\x = – {42^0} + k{.360^0}\end{array} \right.\left( {k \in Z} \right)\\d){\sin ^2}x + {\sin ^2}3x = 1\\ \Rightarrow {\sin ^2}x + {\sin ^2}3x = {\sin ^2}x + {\cos ^2}x\\ \Rightarrow {\sin ^2}3x = {\cos ^2}x\\ \Rightarrow \left[ \begin{array}{l}\sin 3x = \cos x\\\sin 3x = – \cos x\end{array} \right. \Rightarrow \left[ \begin{array}{l}\sin 3x = \sin \left( {\dfrac{\pi }{2} – x} \right)\\\sin 3x = \sin \left( {x – \dfrac{\pi }{2}} \right)\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}3x = \dfrac{\pi }{2} – x + k2\pi \\3x = \pi – \dfrac{\pi }{2} + x + k2\pi \\3x = x – \dfrac{\pi }{2} + k2\pi \\3x = \pi – x + \dfrac{\pi }{2} + k2\pi \end{array} \right. \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\x = \dfrac{\pi }{4} + k\pi \\x = – \dfrac{\pi }{4} + k\pi \\x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{4}\\x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}\end{array} \right.\left( {k \in Z} \right)\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\sin \left( {{{15}^0} – 2x} \right) = – 1\\
\Rightarrow {15^0} – 2x = – {90^0} + k{.360^0}\\
\Rightarrow 2x = {105^0} – k{.360^0}\\
\Rightarrow x = 52,{5^0} + k{.180^0}\\
b)\sin 2x + \sin 5x = 0\\
\Rightarrow sin2x = – sin5x\\
\Rightarrow sin2x = \sin \left( { – 5x} \right)\\
\Rightarrow \left[ \begin{array}{l}
2x = – 5x + k2\pi \\
2x = \pi + 5x + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{7}\\
x = \dfrac{{ – \pi }}{3} – \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
c)\cos \left( {x – {3^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Rightarrow \cos \left( {x – {3^0}} \right) = \cos {45^0}\\
\Rightarrow \left[ \begin{array}{l}
x – {3^0} = {45^0} + k{.360^0}\\
x – {3^0} = – {45^0} + k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = {48^0} + k{.360^0}\\
x = – {42^0} + k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
d){\sin ^2}x + {\sin ^2}3x = 1\\
\Rightarrow {\sin ^2}x + {\sin ^2}3x = {\sin ^2}x + {\cos ^2}x\\
\Rightarrow {\sin ^2}3x = {\cos ^2}x\\
\Rightarrow \left[ \begin{array}{l}
\sin 3x = \cos x\\
\sin 3x = – \cos x
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sin 3x = \sin \left( {\dfrac{\pi }{2} – x} \right)\\
\sin 3x = \sin \left( {x – \dfrac{\pi }{2}} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} – x + k2\pi \\
3x = \pi – \dfrac{\pi }{2} + x + k2\pi \\
3x = x – \dfrac{\pi }{2} + k2\pi \\
3x = \pi – x + \dfrac{\pi }{2} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi \\
x = – \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$