Đáp án:
Câu 6: A
Câu 7: B
Câu 8: D
Giải thích các bước giải:
Câu 6:
Ta có:
$\cos x-\sin x=0$
$\to \cos x=\sin x$
$\to \cos x=\cos(\dfrac{\pi}2-x)$
$\to x=\pm(\dfrac{\pi}2-x)+k2\pi$
Giải $x=(\dfrac{\pi}2-x)+k2\pi$
$\to x=\dfrac{\pi}4+k\pi$
Giải $x=-(\dfrac{\pi}2-x)+k2\pi$
$\to \dfrac{\pi}2=k2\pi$ vô nghiệm
$\to A$
Câu 7:
Ta có:
$\cos(x+\dfrac{\pi}4)=\dfrac12$
$\to x+\dfrac{\pi}4=\pm\dfrac{\pi}3+k2\pi$
$\to x\in\{\dfrac{\pi}{12}+k2\pi, \dfrac{-7\pi}{12}+k2\pi\}$
Ta có: $x\in(-\pi,\pi)$
$\to -\pi<\dfrac{\pi}{12}+k2\pi<\pi$
$\to k=0$ vì $k\in Z$
$\to x=\dfrac{\pi}{12}$
Lại có:
$\to -\pi<\dfrac{-7\pi}{12}+k2\pi<\pi$
$\to k=0$ vì $k\in Z$
$\to x\in\{-\dfrac{31\pi}{12}, \dfrac{-7\pi}{12}\}$
$\to$Tổng nghiệm thỏa mãn đề là:
$\dfrac{\pi}{12}+(\dfrac{-7\pi}{12})=-\dfrac{\pi}2$
$\to B$
Câu 8:
Ta có:
$\sin x\cos\dfrac{\pi}8+\sin\dfrac{\pi}8\cos x=\dfrac12$
$\to \sin(x+\dfrac{\pi}8)=\dfrac12$
$\to x+\dfrac{\pi}8=\dfrac{\pi}6+k2\pi$
$\to x=\dfrac{\pi}{24}+k2\pi$
Do $x\in[-\pi,\pi]$
$\to -\pi\le \dfrac{\pi}{24}+k2\pi\le \pi$
$\to k=0$ vì $k\in Z$
$\to x=\dfrac{\pi}{24}$
Hoặc $x+\dfrac{\pi}8=\pi-\dfrac{\pi}6+k2\pi$
$\to x=\dfrac{17\pi}{24}+k2\pi$
Do $x\in[-\pi,\pi]$
$\to -\pi\le \dfrac{17\pi}{24}+k2\pi\le \pi$
$\to k=0$ vì $k\in Z$
$\to x=\dfrac{17\pi}{24}$
$\to$Tổng nghiệm là:
$\dfrac{\pi}{24}+\dfrac{17\pi}{24}=\dfrac34\pi$
$\to D$
