Chủ đề này có 2 trả lời

[wtuan159]

$sin3x +(\sqrt{3}-2)cos3x=1$

 

[Phạm Hữu Bảo Chung]

Giải

Ta thấy:

$\cos{\dfrac{\pi}{6}} = \dfrac{\sqrt{3}}{2} \Rightarrow 2\cos^2{\dfrac{\pi}{12}} - 1 = \dfrac{\sqrt{3}}{2}$

 

$\Rightarrow \cos{\dfrac{\pi}{12}} = \dfrac{\sqrt{\sqrt{3} + 2}}{2} = \dfrac{\sqrt{4 + 2\sqrt{3}}}{2\sqrt{2}} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}(\sqrt{3} - 1)}$

$\Rightarrow \sin{\dfrac{\pi}{12}} = \dfrac{1}{\sqrt{2}(\sqrt{3} + 1)}$

Phương trình ban đầu tương đương:
$\dfrac{1}{\sqrt{1^2 + (\sqrt{3} - 2)^2}}\sin{3x} + \dfrac{\sqrt{3} - 2}{\sqrt{1^2 + (\sqrt{3} - 2)^2}}\cos{3x} = \dfrac{1}{\sqrt{1^2 + (\sqrt{3} - 2)^2}}$

 

$\Leftrightarrow \dfrac{1}{\sqrt{2}(\sqrt{3} - 1)}\sin{3x} + \dfrac{\sqrt{3} - 2}{\sqrt{2}(\sqrt{3} - 1)}\cos{3x}= \dfrac{1}{\sqrt{2}(\sqrt{3} - 1)}$

$\Leftrightarrow \cos{\dfrac{\pi}{12}}\sin{3x} - \sin{\dfrac{\pi}{12}}\cos{3x} = \cos{\dfrac{\pi}{12}}$

$\Leftrightarrow \sin{\left ( 3x - \dfrac{\pi}{12}\right )} = \sin{\dfrac{5\pi}{12}}$

Bạn tự giải tiếp nhé!

[Messi10597]

PT$\Leftrightarrow (\sqrt{3}-2)cos3x=1-sin3x$

    $\Leftrightarrow (\sqrt{3}-2)(cos\frac{3x}{2}-sin\frac{3x}{2})(cos\frac{3x}{2}+sin\frac{3x}{2})=(cos\frac{3x}{2}-sin\frac{3x}{2})^{2}$

    $\Leftrightarrow (cos\frac{3x}{2}-sin\frac{3x}{2})[(\sqrt{3}-3)cos\frac{3x}{2}+(\sqrt{3}-1)sin\frac{3x}{2}]=0$

 đến đây dễ rồi