Đáp án:
Giải thích các bước giải:
1) `2sin(x+\frac{\pi}{4})+sin(x-\frac{\pi}{4})=\frac{3\sqrt{2}}{2}`
`⇔ 2\frac{sin\ x+cos\ x}{\sqrt{2}}+\frac{sin\ x-cos\ x}{\sqrt{2}}=\frac{3\sqrt{2}}{2}`
`⇔ 3sin\ x+cos\ x=3`
`⇔ \frac{3}{\sqrt{10}}sin\ x+\frac{1}{\sqrt{10}}cos\ x=\frac{3}{\sqrt{10}}`
Đặt `\frac{3}{\sqrt{10}}=cos\ \alpha,\frac{1}{\sqrt{10}}=sin\ \alpha`
`⇔ sin\ x.cos\ \alpha+cos\ x.sin\ \alpha=\frac{3}{\sqrt{10}}`
`⇔ sin (x+\alpha)=sin(\frac{\pi}{2}-\alpha)`
`⇔` \(\left[ \begin{array}{l}x+\alpha=\dfrac{\pi}{2}-\alpha+k2\pi\ (k \in \mathbb{Z})\\x+\alpha=\pi-\dfrac{\pi}{2}+\alpha+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}-2\alpha+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{2}-2\alpha+k2\pi\ (k \in \mathbb{Z});\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})}`
2) `\sqrt{3}cos\ 2x+sin\ 2x+2sin(2x-\frac{\pi}{6})=2\sqrt{2}`
`⇔ \frac{\sqrt{3}}{2}cos\ 2x+\frac{1}{2}sin\ 2x+sin(2x-\frac{\pi}{6})=\sqrt{2}`
`⇔ cos\ \frac{\pi}{6}.cos\ 2x+sin\ \frac{\pi}{6}.sin\ 2x+sin(2x-\frac{\pi}{6})=\sqrt{2}`
`⇔ cos (2x-\frac{\pi}{6})+sin(2x-\frac{\pi}{6})=\sqrt{2}`
`⇔ \sqrt{2}sin(2x-\frac{\pi}{6}+\frac{\pi}{4})=\sqrt{2}`
`⇔ sin (2x+\frac{\pi}{12})=1`
`⇔ 2x+\frac{\pi}{12}=\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})`
`⇔ 2x=\frac{5\pi}{12}+k2\pi\ (k \in \mathbb{Z})`
`⇔ x=\frac{5\pi}{24}+k\pi\ (k \in \mathbb{Z})`
Vậy `S={\frac{5\pi}{24}+k\pi\ (k \in \mathbb{Z})}`
