1. $\frac{(1+sinx+cos2x)sin(x+\frac{\Pi }{4})}{1+tanx}=\frac{1}{\sqrt{2}}cosx$
2. $\frac{(1-2sinx)cosx}{(1+2sinx)(1-sinx)}=\sqrt{3}$
1. $\frac{(1+sinx+cos2x)sin(x+\frac{\Pi }{4})}{1+tanx}=\frac{1}{\sqrt{2}}cosx$
2. $\frac{(1-2sinx)cosx}{(1+2sinx)(1-sinx)}=\sqrt{3}$
1. $\frac{(1+sinx+cos2x)sin(x+\frac{\Pi }{4})}{1+tanx}=\frac{1}{\sqrt{2}}cosx$
2. $\frac{(1-2sinx)cosx}{(1+2sinx)(1-sinx)}=\sqrt{3}$
1. ĐK: $x\neq {\frac{\Pi }{2}+k\Pi; -\frac{\Pi}{4}+k\Pi}$
PT $\Leftrightarrow (1+sin x +cos 2x).\frac{sin x+ cos x}{\sqrt{2}}=\frac{1}{\sqrt{2}}cos x(1+\frac{sin}{cosx})\Leftrightarrow (sin x+cos x)(sin x+cos 2x)=0\Leftrightarrow \begin{bmatrix} sin x=-cos x ( tan x=-1 ,L) & \\ sin x= cos 2x=1-2sin^2 x& \end{bmatrix}\Leftrightarrow \begin{bmatrix} sin x=1 (cos x=0, L) & \\ sin x= \frac{-1}{2}& \end{bmatrix}\Leftrightarrow \begin{bmatrix} x=-\frac{\pi}{6}+k2\pi & \\ x=-\frac{5\pi}{6}+k2\pi& \end{bmatrix}$
2. ĐK: $\left\{\begin{matrix} sin x\neq\frac{-1}{2} & \\ sin x \neq 1& \end{matrix}\right. $
PT$\Leftrightarrow cos x-sin 2x= \sqrt{3}(1+sin x-2sin^2x)\Leftrightarrow cos x-sin 2x=\sqrt{3}(sin x+cos2x)\Leftrightarrow \frac{1}{2}cosx-\frac{\sqrt{3}}{2}sin x=\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sin x\Leftrightarrow cos(x-\pi /3)=cos(2x-\pi/6)\Leftrightarrow \begin{bmatrix} x-\pi /3=2x-\pi/6+k2\pi& \\ x-\pi /3=-2x+\pi/6+k2\pi& \end{bmatrix}$
Bài viết đã được chỉnh sửa nội dung bởi Truong Gia Bao: 09-07-2017 - 22:16
0 thành viên, 0 khách, 0 thành viên ẩn danh