Giải các pt sau: a) sin(πsinx) =1 b) tan(π/4cosx) =1 Giúp em vs ạ đang cần gấp ạ 08/09/2021 Bởi Jasmine Giải các pt sau: a) sin(πsinx) =1 b) tan(π/4cosx) =1 Giúp em vs ạ đang cần gấp ạ
Đáp án: $\begin{array}{l}a)\sin \left( {\pi \sin x} \right) = 1\\ \Rightarrow \pi .\sin x = \dfrac{\pi }{2} + k2\pi \\ \Rightarrow \sin x = \dfrac{1}{2} + 2k\\Do: – 1 \le \sin x \le 1\\ \Rightarrow k = 0\\ \Rightarrow \sin x = \dfrac{1}{2}\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \pi – \dfrac{\pi }{6} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\b)tan\left( {\dfrac{\pi }{4}.\cos x} \right) = 1\\ \Rightarrow \dfrac{\pi }{4}.\cos x = \dfrac{\pi }{4} + k\pi \\ \Rightarrow \cos x = 1 + \dfrac{k}{4}\\Do: – 1 \le \cos x \le 1\\ \Rightarrow k = 0\\ \Rightarrow \cos x = 1\\ \Rightarrow x = k2\pi \end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\sin \left( {\pi \sin x} \right) = 1\\
\Rightarrow \pi .\sin x = \dfrac{\pi }{2} + k2\pi \\
\Rightarrow \sin x = \dfrac{1}{2} + 2k\\
Do: – 1 \le \sin x \le 1\\
\Rightarrow k = 0\\
\Rightarrow \sin x = \dfrac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi – \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
b)tan\left( {\dfrac{\pi }{4}.\cos x} \right) = 1\\
\Rightarrow \dfrac{\pi }{4}.\cos x = \dfrac{\pi }{4} + k\pi \\
\Rightarrow \cos x = 1 + \dfrac{k}{4}\\
Do: – 1 \le \cos x \le 1\\
\Rightarrow k = 0\\
\Rightarrow \cos x = 1\\
\Rightarrow x = k2\pi
\end{array}$