Giải phương trình $\text{co}\text{s}^4x+\text{si}\text{n}^4x+c\text{os}\left({x-\dfrac\pi 4}\right)\sin \left({3x-\dfrac\pi 4}\right)-\dfrac32=0.$
A.
$x=\dfrac\pi 4+k2\pi ,k\in \mathbb{Z}$.
B.
$x=\dfrac\pi 4+k\pi ,k\in \mathbb{Z}$.
C.
$x=\dfrac\pi 4+k\dfrac\pi 2,k\in \mathbb{Z}$.
D.
$x=-\dfrac\pi 4+k\pi ,k\in \mathbb{Z}$.
Đáp án và lời giải
Đáp án:B
Lời giải:$\text{co}\text{s}^4x+\text{si}\text{n}^4x+c\text{os}\left({x-\dfrac\pi 4}\right)\sin \left({3x-\dfrac\pi 4}\right)-\dfrac32=0$ $\Leftrightarrow 1-2\text{si}\text{n}^2x.\cos ^2x+\dfrac12\left[{\sin \left({4x-\dfrac\pi 2}\right)+\sin 2x}\right]-\dfrac32=0$ $\Leftrightarrow \text{si}\text{n}^22x+\sin 2x-2=0$ $\Leftrightarrow \left[\begin{aligned}& \sin 2x=1\Leftrightarrow x=\dfrac\pi 4+k\pi ,k\in \mathbb{Z} \\ & \sin 2x=-2\text{ }(ktm) \end{aligned}\right..$