Giải pt
$ a, \cos2x + \sqrt{3}  \sin2x = 2 \cos3x$
$b, \sin2x - \cos 2x = \sqrt{2} \sin(x+\pi/6)$
$c, \sin^2x - \sin2x - 3\cos^2x =0$
$d, 3\sin^2x - \sin 2x + \cos^2x = 2$

Bài 2 : tìm max min của h.s
$a, y = 2\cos2x - \sin x$
$b, y= 3\cos2x - 2\sin2x$
$c, y = \sin^2x - \sin2x + 3cos^2x$
Câu b áp dụng công thức $sin 2x-cos 2x=\sqrt{2}sin( 2x-\pi/4)$
Câu c,d cùng 1 dạng: Đầu tiên bạn xét cos x=0 có phải là nghiệm k
Nếu không bạn chia cả 2 vế cho $cos^{2} x$ và áp dụng các công thức sau $\frac{sin x}{cos x}=tan x$ $\frac{1}{cos^{2} x}=1+tan^{2} x$ để đưa pt về 1 ẩn theo hàm tan.Nếu mà cos x=0 tm thi bạn kl đó là nghiệm pt và xét TH $cos x\neq0$ để chia
a) Bạn chia cả 2 vế pt cho 2 là ra $\cos (2x-\pi/3)=cos 3x$

Bạn cần đăng nhập để có thể gửi đáp án

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