1. $cos^{3}x + cos^{2}x + 2sinx - 2 = 0$
2. $2+(2+sin2x)(\frac{1}{sinx}+\frac{1}{cosx}+tanx+cotx)=0$
3. $\frac{1}{sinxcosx}=2sin2x+2cos2x$
1. $cos^{3}x + cos^{2}x + 2sinx - 2 = 0$
2. $2+(2+sin2x)(\frac{1}{sinx}+\frac{1}{cosx}+tanx+cotx)=0$
3. $\frac{1}{sinxcosx}=2sin2x+2cos2x$
1. Ta có:
$\cos^3{x}+\cos^2{x}+2\sin{x}-2=0 \\ \Leftrightarrow \cos^2(1+\cos{x}-2(1-\sin{x})=0 \\ \Leftrightarrow (1-\sin{x})(\sin{x}\cos{x}+\sin{x}+\cos{x}-1)=0$
$\Leftrightarrow x=\frac{\pi}{2}+2k\pi (k \in \mathbb{Z}) \vee \sin{x}\cos{x}+\sin{x}+\cos{x}-1=0$
Đặt $\sin{x}\cos{x}=P, \sin{x}+\cos{x}=S$, ta có hệ:
$\begin{cases}S+P-1=0 \\ S^2-2P=1\end{cases}$
Bài viết đã được chỉnh sửa nội dung bởi Arthur Pendragon: 06-08-2019 - 20:24
"WHEN YOU HAVE ELIMINATED THE IMPOSSIBLE, WHATEVER REMAINS, HOWEVER IMPROBABLE, MUST BE THE TRUTH"
-SHERLOCK HOLMES-
2. ĐK: $x \neq k\frac{\pi}{2}$
Ta có:
$ 2+(2+\sin{2x})\left(\frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{1}{\sin{x}\cos{x}}\right)+\sin{2x}\left(\frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{1}{\sin{x}\cos{x}}\right)-2(\sin{x}+\cos{x})=0$
$\Leftrightarrow 2(\sin{x}+\cos{x})\left[(\sin{x}+\cos{x})\left(\frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{1}{\sin{x}\cos{x}}\right)-2\right]=0$
$\Leftrightarrow (\sin{x}+cos{x})\left(\frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{1}{\sin{x}\cos{x}}\right)-2=0 $ (do $\sin{x}+\cos{x} \neq 0$)
$\Leftrightarrow \frac{1}{\sin{x}}+\frac{1}{\cos{x}}+\frac{1}{\sin{x}\cos{x}}=0$
Bài viết đã được chỉnh sửa nội dung bởi Arthur Pendragon: 06-08-2019 - 20:55
"WHEN YOU HAVE ELIMINATED THE IMPOSSIBLE, WHATEVER REMAINS, HOWEVER IMPROBABLE, MUST BE THE TRUTH"
-SHERLOCK HOLMES-
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