3 sin^2 x – 5 Sin x – 2 = 0 3cos^2 x- sin^2 x – sin2x = 0 28/07/2021 Bởi aikhanh 3 sin^2 x – 5 Sin x – 2 = 0 3cos^2 x- sin^2 x – sin2x = 0
Đáp án: $\begin{array}{l} + )3{\sin ^2}x – 5\sin x – 2 = 0\\ \Rightarrow 3{\sin ^2}x – 6\sin x + \sin x – 2 = 0\\ \Rightarrow \left( {\sin \,x – 2} \right)\left( {3\sin \,x + 1} \right) = 0\\ \Rightarrow \sin \,x = – \frac{1}{3}\left( {do:\sin \,x \le 1} \right)\\ \Rightarrow \left[ \begin{array}{l}x = \arcsin \left( { – \frac{1}{3}} \right) + k2\pi \\x = \pi – \arcsin \left( { – \frac{1}{3}} \right) + k2\pi \end{array} \right.\\ + )3{\cos ^2}x – {\sin ^2}x – \sin 2x = 0\\ \Rightarrow 3{\cos ^2}x – 2\sin x.\cos x – {\sin ^2}x = 0\\ \Rightarrow 3{\cos ^2}x – 3\sin x.\cos x + \sin x.\cos x – {\sin ^2}x = 0\\ \Rightarrow \left( {\cos x – \sin x} \right)\left( {3\cos x + \sin x} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}\cos x = \sin x\\3\cos x = – \sin x\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x = – 3\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = \arctan \left( { – 3} \right) + k\pi \end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
+ )3{\sin ^2}x – 5\sin x – 2 = 0\\
\Rightarrow 3{\sin ^2}x – 6\sin x + \sin x – 2 = 0\\
\Rightarrow \left( {\sin \,x – 2} \right)\left( {3\sin \,x + 1} \right) = 0\\
\Rightarrow \sin \,x = – \frac{1}{3}\left( {do:\sin \,x \le 1} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \arcsin \left( { – \frac{1}{3}} \right) + k2\pi \\
x = \pi – \arcsin \left( { – \frac{1}{3}} \right) + k2\pi
\end{array} \right.\\
+ )3{\cos ^2}x – {\sin ^2}x – \sin 2x = 0\\
\Rightarrow 3{\cos ^2}x – 2\sin x.\cos x – {\sin ^2}x = 0\\
\Rightarrow 3{\cos ^2}x – 3\sin x.\cos x + \sin x.\cos x – {\sin ^2}x = 0\\
\Rightarrow \left( {\cos x – \sin x} \right)\left( {3\cos x + \sin x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \sin x\\
3\cos x = – \sin x
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = – 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k\pi \\
x = \arctan \left( { – 3} \right) + k\pi
\end{array} \right.
\end{array}$
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