2(cosx+can3sinx)+5(sinx+can3cosx+1)=0 giup minh voi a!!!

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1. Đáp án:

    

   Giải thích các bước giải:

   $\quad 2\left(\cos x + \sqrt3\sin x\right) + 5\left(\sin x + \sqrt3\cos x +1\right)= 0$

   $\Leftrightarrow \left(5+2\sqrt3\right)\sin x + \left(2 + 5\sqrt3\right)\cos x + 5 = 0$

   $\Leftrightarrow \dfrac{5 + 2\sqrt3}{2\sqrt{29 + 10\sqrt3}}\sin x + \dfrac{2+ 5\sqrt3}{2\sqrt{29 + 10\sqrt3}}\cos x = -\dfrac{5}{2\sqrt{29 + 10\sqrt3}}$

   Đặt $\begin{cases}\cos\alpha = \dfrac{5 + 2\sqrt3}{2\sqrt{29 + 10\sqrt3}}\\\sin\alpha = \dfrac{2+ 5\sqrt3}{2\sqrt{29 + 10\sqrt3}}\end{cases}\Rightarrow \alpha = \arccos\left(\dfrac{5 + 2\sqrt3}{2\sqrt{29 + 10\sqrt3}}\right)$

   Phương trình trở thành:

   $\quad \sin x\cos\alpha + \cos x\sin\alpha = -\dfrac{5}{2\sqrt{29 + 10\sqrt3}}$

   $\Leftrightarrow \sin(x + \alpha)= -\dfrac{5}{2\sqrt{29 + 10\sqrt3}}$

   $\Leftrightarrow \left[\begin{array}{l}x + \alpha = \arcsin\left(-\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\\x + \alpha = \pi – \arcsin\left(-\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = – \alpha -\arcsin\left(\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\\x =\pi -\alpha + \arcsin\left(\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = – \arccos\left(\dfrac{5 + 2\sqrt3}{2\sqrt{29 + 10\sqrt3}}\right) -\arcsin\left(\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\\x =\pi -\arccos\left(\dfrac{5 + 2\sqrt3}{2\sqrt{29 + 10\sqrt3}}\right) + \arcsin\left(\dfrac{5}{2\sqrt{29 + 10\sqrt3}}\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

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