phương trình lượng giác - 11

B

buichianh18896

mình gợi ý

$\begin{array}{l}
2{\cos ^3}x + \cos 2x + \sin x = 0\\
\Leftrightarrow 2{\cos ^3}x + {\cos ^2}x - {\sin ^2}x + \sin x = 0\\
\Leftrightarrow (2\cos x + 1){\cos ^2}x - \sin x(\sin x - 1) = 0\\
\Leftrightarrow (2\cos x + 1)(1 - {\sin ^2}x) - \sin x(\sin x - 1) = 0\\
\Leftrightarrow (1 - \sin x)[(2\cos x + 1)(1 + \sin x) + \sin x] = 0\\
\Leftrightarrow (1 - \sin x)(2\cos x + 2\sin x\cos x + 1 + 2\sin x) = 0\\
\Leftrightarrow (1 - \sin x)[2(\sin x + \cos x) + 2\sin x\cos x + 1] = 0
\end{array}$
$\begin{array}{l}
\Leftrightarrow hoac\_\sin x = 1\_\_\_\_(1)\\
hoac\_[2(\sin x + \cos x) + 2\sin x\cos x + 1] = 0\_\_\_\_(2)\\
pt\_(2):dat:\sin x + \cos x = t, - \sqrt 2 \le t \le \sqrt 2 \\
{t^2} = 1 + 2\sin x\cos x\_\_\_thay\_vao\_(2)
\end{array}$
 
Last edited by a moderator:
N

nguyendat367

buichianh18896 said:
$\begin{array}{l}
2{\cos ^3}x + \cos 2x + \sin x = 0\\
\Leftrightarrow 2{\cos ^3}x + {\cos ^2}x - {\sin ^2}x + \sin x = 0\\
\Leftrightarrow (2\cos x + 1){\cos ^2}x - \sin x(\sin x - 1) = 0\\
\Leftrightarrow (2\cos x + 1)(1 - {\sin ^2}x) - \sin x(\sin x - 1) = 0\\
\Leftrightarrow (1 - \sin x)[(2\cos x + 1)(1 + \sin x) + \sin x] = 0
\end{array}$
Bấm để xem đầy đủ nội dung ...

e k hỉu gì hjt ?? a nhập binh thường di:confused::confused::confused::confused::confused:
 
B

buichianh18896

sao the nhi cong thuc lai the nhi?

.....................................................................................................................................................................................................................
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom