Монгол Бодлогын Сан

$3\cos 2x-4\sin2x>\dfrac{5}{\sqrt{2}}$ $(\alpha=\arccos \dfrac{\fbox{a}}{5}$ туслах өнцөг оруулбал) $\Leftrightarrow \cos(2x+\alpha)>\dfrac{1}{\sqrt{\fbox{b}}}\Leftrightarrow x\in \left]-\dfrac{\pi}{\fbox{c}}-\dfrac{\alpha}{2}+\pi k;\dfrac{\pi}{\fbox{d}}-\dfrac{\alpha}{2}+\pi k\right[ (k\in \mathbb Z).$