Đáp án:
$x=\{\dfrac{3\pi}4;\dfrac{7\pi}4,\dfrac{2\pi}5;\dfrac{7\pi}5\}$
Lời giải:
$4\sin x+6\cos x=\dfrac{1}{\cos x}$
Đk: $\cos x\ne 0\Leftrightarrow x\ne \dfrac{\pi}{2}+k\pi$
Phương trình tương đương:
$4\sin x\cos x+6{\cos}^2x=1$
$\Rightarrow 2\sin 2x+3(\cos 2x+1)=1$
$\Rightarrow 2\sin 2x+3\cos 2x=-2$
$\Rightarrow\dfrac{2}{\sqrt{13}}\sin 2x+\dfrac{3}{\sqrt{13}}\cos 2x=\dfrac{-2}{\sqrt{13}}$
Đặt $\cos \alpha=\dfrac{2}{\sqrt{13}}$ và $\sin\alpha=\dfrac{3}{\sqrt{13}}$
Phương trình tương đương:
$\cos\alpha\sin2x+\sin\alpha\cos2x=\dfrac{-2}{\sqrt{13}}$
$\Rightarrow \sin(2x+\alpha)=\dfrac{-2}{\sqrt{13}}$
$\Rightarrow \left[ \begin{array}{l} 2x+\alpha=\arcsin\dfrac{-2}{\sqrt{13}}+k2\pi \\ 2x+\alpha=\pi-\arcsin\dfrac{-2}{\sqrt{13}}+k2\pi \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{-\alpha}{2}+\dfrac{1}{2}\arcsin\dfrac{-2}{\sqrt{13}}+k\pi=-\dfrac{\pi}4+k\pi \\ x=\dfrac{-\alpha}{2}+\dfrac{\pi}{2}-\dfrac{1}{2}\arcsin\dfrac{-2}{\sqrt{13}}+k\pi≈\dfrac{2\pi}5+k\pi\end{array} \right .(k\in\mathbb Z)$
$x\in(0;2\pi)$
$\Rightarrow x=\{\dfrac{3\pi}4;\dfrac{7\pi}4,\dfrac{2\pi}5;\dfrac{7\pi}5\}$.
