cos 2x +3 sin x-2=0 (Giải PT tiết hộ mình nhe ) ^***^ 11/08/2021 Bởi Rose cos 2x +3 sin x-2=0 (Giải PT tiết hộ mình nhe ) ^***^
Đáp án: $\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$ Giải thích các bước giải: $cos2x + 3sinx – 2 = 0$ $\Leftrightarrow 1 – 2sin^2x + 3sinx – 2 = 0$ $\Leftrightarrow 2sin^2x – 3sinx +1 = 0$ $\Leftrightarrow \left[\begin{array}{l}sinx = 1\\sinx =\dfrac{1}{2}\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$ Bình luận
Đáp án: \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\) Giải thích các bước giải: \(\cos 2x +3\sin x-2=0\) \(\Leftrightarrow 1-2\sin^{2} x+3\sin x-2=0\) \(\Leftrightarrow -2\sin^{2} x +3\sin x-1=0\) \((-1 \leq \sin x \leq 1)\) \(\Leftrightarrow \) \(\left[ \begin{array}{l}\sin x=1\\\sin x=\dfrac{1}{2}\end{array} \right.\) \(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\sin x=\sin \dfrac{\pi}{6}\end{array} \right.\) \(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\) Bình luận
Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$cos2x + 3sinx – 2 = 0$
$\Leftrightarrow 1 – 2sin^2x + 3sinx – 2 = 0$
$\Leftrightarrow 2sin^2x – 3sinx +1 = 0$
$\Leftrightarrow \left[\begin{array}{l}sinx = 1\\sinx =\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$
Đáp án:
\(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\)
Giải thích các bước giải:
\(\cos 2x +3\sin x-2=0\)
\(\Leftrightarrow 1-2\sin^{2} x+3\sin x-2=0\)
\(\Leftrightarrow -2\sin^{2} x +3\sin x-1=0\) \((-1 \leq \sin x \leq 1)\)
\(\Leftrightarrow \) \(\left[ \begin{array}{l}\sin x=1\\\sin x=\dfrac{1}{2}\end{array} \right.\)
\(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\sin x=\sin \dfrac{\pi}{6}\end{array} \right.\)
\(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\)