Đáp án:

Giải thích các bước giải:

$a)\quad \sin x - \cos x = \dfrac{\sqrt6}{2}$

$\Leftrightarrow \sqrt2\sin\left(x -\dfrac{\pi}{4}\right)= \dfrac{\sqrt6}{2}$

$\Leftrightarrow \sin\left(x -\dfrac{\pi}{4}\right)= \dfrac{\sqrt3}{2}$

$\Leftrightarrow \sin\left(x -\dfrac{\pi}{4}\right)= \sin\dfrac{\pi}{3}$

$\Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{4}=\dfrac{\pi}{3} + k2\pi\\x -\dfrac{\pi}{4} = \dfrac{2\pi}{3} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{7\pi}{12} + k2\pi\\x= \dfrac{11\pi}{12} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

$c)\quad \sin4x +\cos4x = \sqrt3$

$\Leftrightarrow \sqrt2\sin\left(4x + \dfrac{\pi}{4}\right)= \sqrt3$

$\Leftrightarrow \sin\left(4x + \dfrac{\pi}{4}\right)=\dfrac{\sqrt6}{2}$ (vô nghiệm)

Vậy phương trình đã cho vô nghiệm

$e)\quad 3\sin x + \sqrt3\cos x= 1$

$\Leftrightarrow \dfrac{\sqrt3}{2}\sin x + \dfrac12\cos x = \dfrac{\sqrt3}{6}$

$\Leftrightarrow \sin\left(x + \dfrac{\pi}{6}\right)= \dfrac{\sqrt3}{6}$

$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{6}=\arcsin\dfrac{\sqrt3}{6} + k2\pi\\x +\dfrac{\pi}{6} = \pi -\arcsin\dfrac{\sqrt3}{6} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{6}+\arcsin\dfrac{\sqrt3}{6} + k2\pi\\x = \dfrac{5\pi}{6} -\arcsin\dfrac{\sqrt3}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

`~rai~`

\(a)sinx-cosx=\dfrac{\sqrt{6}}{2}\\\Leftrightarrow \sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\\\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\\\Leftrightarrow \left[\begin{array}{I}x-\dfrac{\pi}{4}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{7\pi}{12}+k2\pi\\x=\dfrac{11\pi}{12}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\c)sin4x+cos4x=\sqrt{3}\\\Leftrightarrow \sqrt{2}sin\left(4x+\dfrac{\pi}{4}\right)=\sqrt{3}\\\Leftrightarrow sin\left(4x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\text{(vô nghiệm do}\dfrac{\sqrt{6}}{2}>1)\\e)3sinx+\sqrt{3}cosx=1\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\\\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{6}\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{6}=arcsin\dfrac{\sqrt{3}}{2}+k2\pi\\x+\dfrac{\pi}{6}=\pi-arcsin\dfrac{\sqrt{3}}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{6}+arcsin\dfrac{\sqrt{3}}{6}+k2\pi\\x=\dfrac{5\pi}{6}-arcsin\dfrac{\sqrt{3}}{6}+k2\pi\end{array}\right.\quad(k\in\mathbb{Z})\)