Chủ đề này có 1 trả lời

[Hoang_kang]

$\sqrt{2}.\cos\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)- \sqrt{6}.\sin\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)=2\sin\left(\dfrac{x}{5}- 2\dfrac{\pi}{3}\right)-2\sin\left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

Mod: xem code:

\sqrt{2}.\cos\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)- \sqrt{6}.\sin\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)=2\sin\left(\dfrac{x}{5}- 2\dfrac{\pi}{3}\right)-2\sin\left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)

Bài viết đã được chỉnh sửa nội dung bởi h.vuong_pdl: 21-06-2011 - 07:02

[h.vuong_pdl]

$\sqrt{2}.\cos\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)- \sqrt{6}.\sin\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)=2\sin\left(\dfrac{x}{5}- 2\dfrac{\pi}{3}\right)-2\sin\left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

Mod: xem code:

@@@: Note:

$\sqrt{2}.\cos\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right)- \sqrt{6}.\sin\left(\dfrac{x}{5}- \dfrac{\pi}{12}\right) = 2\sqrt{2}.\cos\left(\dfrac{x}{5}+\dfrac{\pi}{4}\right)=2\cos\dfrac{x}{5} - 2\sin\dfrac{x}{5}$

$2\sin\left(\dfrac{x}{5}- 2\dfrac{\pi}{3}\right) = -\sin\dfrac{x}{5} - \sqrt{3}\cos\dfrac{x}{5}, \\.\\ 2\sin\left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right) = \sqrt{3}\sin\dfrac{3x}{5}+\cos\dfrac{3x}{5}$

Suy ra:

$\textup{pt} \Leftrightarrow 2\cos y -2\sin y = -\sin y -\sqrt{3}\cos y - \sqrt{3}\sin3y - \cos3y \\.\\ \Leftrightarrow \cos y\Left(2+\sqrt{3} + (4\cos^2x-3)\right) = \sin y\left(1-\sqrt{3}(3-4\sin^2x)\right)$

Dat $t = \cot y = \dfrac{\cos y}{\sin y}$, ta dua ve giai phuong trinh bac 3 an t.

Bài viết đã được chỉnh sửa nội dung bởi h.vuong_pdl: 21-06-2011 - 07:37