\(\Leftrightarrow2cos^2x-1+9cosx+4=0\)
\(\Leftrightarrow2cos^2x+9cosx+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{-9-\sqrt{57}}{4}\left(l\right)\\cosx=\dfrac{-9+\sqrt{57}}{4}\left(nh\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\alpha+k2\pi\)
\(\frac{cos^2X-2cos\left(X+\frac{3Π}{4}\right)Sin\left(3x-\frac{Π}{4}\right)-2}{2cosx-\sqrt{2}}=0\)
1. 2cos2x + sinx = sin3x
2. cos2x + 2(sin3x-1)sin2(π/4 - x/2) = 0
\(1+\sqrt{2}Sin\left(X+\frac{Π}{4}\right)+sin2x+cos2x=0\)
Tìm đồng biến nghịch biến của y = sin2x trên x ∈ (0; \(^{\frac{\text{π}}{2}}\))
Tìm đồng biến nghịch biến của y = cos2x trên x ∈ (-\(\frac{\text{π}}{4}\); \(\frac{\text{π}}{4}\))
Bài 1:
a) 4sin23x + 2(\(\sqrt{3}\)+1) cos 3x - \(\sqrt{3}\)= 4
b) cos2x + 9cosx + 5 = 0
c) 4cos5(2 - 6x) + 16cos2(1 - 3x) =13
d)\(\frac{1}{cos^2x}-\left(3+\sqrt{3}\right)tanx-3+\sqrt{3}=0\)
e) \(\frac{3}{cosx}+tan^2x=9\)
f) 9 - 13cosx + \(\frac{4}{1+tan^2x}=0\)
g) \(\frac{1}{sin^2x}=cotx+3\)
h) \(\frac{1}{cos^2x}+3cot^2x=5\)
i) cos2x - 3cosx = 4cos2\(\frac{x}{2}\)
k) 2cos2x + tanx=\(\frac{4}{5}\)
cho pt; 1-sin2x+cos2x-2mcosx=0
tìm m để pt có 2 nghiệm thuộc (-π;π)
giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1) cos3x - cos4x + cos5x =0
2) sin3x + cos2x = 1 + 2sinx.cos2x
3) cos2x - cosx = 2 sin\(^2\)\(\dfrac{3x}{2}\)
4) cos\(^2\)2x + cos\(^2\)3x = sin\(^2\)x
5) sin3x.sin5x - cos4x.cos6x = 0
A) 3Cos4x-Sin2x+Cos2x-2=0
B) 1/Sin2x+3Cotx+1=0
C) Cos2x-3Cosx=4Cos2x/2
