Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC
Các câu hỏi tương tự
Tìm GTLN, GTNN:
a, \(y=4\sin^2x-4\sin x+3\).
b, \(y=\cos^2x+2\sin x+2\).
c, \(y=\sin^4x-2\cos^2x+1\).
1 tháng 8 2020 lúc 12:20
giải các pt
a) \(cos^2x+sin2x-1=0\)
b) \(\sqrt{3}sin2x+\:cos^4x-sin^4x=\sqrt{2}\)
c) \(\:cos^2x-sin^2x=\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\)
d) \(4\left(sin^4x+cos^4x\right)+\sqrt{3}.sin4x=2\)
e) \(4sinx.cosx.cos2x+cos4x=\sqrt{2}\)
a)sin^4\(\frac{x}{3}\) +cos^4\(\frac{x}{3}\)=\(\frac{5}{8}\)
b)4(sin^4x+cos^4x)+\(\sqrt{3}\)sin4x=2
c)cos^4x+sin^6x=cos2x
d)cos^6x+sin^6x=cos4x
2cos^2x+2cos^2x+4cos^3(2x)-3cos2x=5
1. Tìm m để PT có nghiệm:
a) \(\sqrt{3}\cos^2x+\dfrac{1}{2}\sin2x=m\)
b) \(3\sin^2x-2\sin x\cos x+m=0\)
c) \(\sin^2x+2\left(m-1\right)\sin x\cos x-\left(m+1\right)\cos^2x=m\)
1. Tìm m để PT có nghiệm:
a) \(\sqrt{3}\cos^2x+\dfrac{1}{2}\sin2x=m\)
b) \(3\sin^2x-2\sin x\cos x+m=0\)
c) \(^{ }\sin^2x+2\left(m-1\right)\sin x\cos x-\left(m+1\right)\cos^2x=m\)
Tìm TXĐ:
a, \(y=\dfrac{1}{2}\sin\left(2x-1\right)-\cos\left(x^2-2\right)\).
b, \(y=\sin\sqrt{2x-4}\).
c, \(y=\sqrt{1-\cos^2x}\).
Giải PT
a) \(\left|\sin x-\cos x\right|+\left|\sin x+\cos x\right|=2\)
b) \(\tan x-3\cot x=4\left(\sin x+\sqrt{3}\cos x\right)\)
c) \(2\sin^2x-2\sqrt{2}\sin x+3\tan^22x-2\sqrt{3}\tan2x+2=0\)
a) cos^6x+sin^2x=1
b)cos^6x-sin^6x=13/18cos^2(2x)
c)cos^4x+sin^6x=cos2x
d)2cos^2(2x)+cos2x=4sin^2(2x) cos^2x
giải pt
a, \(\sin^2x+\sin^22x+\sin^23x=\dfrac{3}{2}\)
b. \(\cos^2x+\sin^22x+\cos^23x=1\)
c,\(\sin5x+2\cos^2x=1\)
d,\(1+\tan x=2\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\)
e,\(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)