Đặt \(cosx=t\) với \(-1\le t\le1\)
\(\Rightarrow4t^2+2t-2-\sqrt{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{-1+\sqrt{\left(2\sqrt{2}+1\right)^2}}{4}=\frac{\sqrt{2}}{2}\\t=\frac{-1-\sqrt{\left(2\sqrt{2}+1\right)^2}}{4}=\frac{-1-\sqrt{2}}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cosx=\frac{\sqrt{2}}{2}\)
\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)
Giai phương trình bậc nhất :
a/ \(2sinx+\sqrt{3}=0\)
b/ \(2cosx-\sqrt{3}=0\)
c/ \(2cosx-\sqrt{2}=0\)
d/ \(tanx+\sqrt{3}=0\)
\(4cos^2x+2cosx-2-\sqrt{2}=0\)
\(\dfrac{2sin^3x+2\sqrt{3}sin^2x.cosx-2sin^2x+cos\left(2x+\dfrac{\pi}{3}\right)}{2cosx-\sqrt{3}}=0\)
\(\frac{cos^2X-2cos\left(X+\frac{3Π}{4}\right)Sin\left(3x-\frac{Π}{4}\right)-2}{2cosx-\sqrt{2}}=0\)
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) sin2x + 2cosx = 0
2) sin(2x -10*) = \(\dfrac{1}{2}\) (-120* <x< 90*)
3) cos(2x+10*)= \(\dfrac{\sqrt{2}}{2}\)(-180*<x<180*)
4) \(\sin^2\left(5x+\dfrac{2\pi}{5}\right)-\cos^2\)(\(\dfrac{x}{4}-\pi\)) =0
giải pt : \(\dfrac{2cos2x+1}{\sqrt{3}sinx+cosx}\)=2cosx-1
tìm txđ hàm số D: y=\(\dfrac{2+3sinx}{2sin2x+\sqrt{2}}\)
Giải pt ( đưa về pt bậc 2 )
1. tan2x - 5tanx + 6 = 0
2. 3cos22x + 4cos2x + 1 = 0
1) 2sinx + cosx = sin2x + 1
2) (1 + cosx)(1+sinx) = 2
3) 3cos4x - 8cos6x + 2cos2x +3 =0
4) sin3x + cos3x.sinx + cosx = \(\sqrt{2}\)cos2x
5) (2cosx -1)(2sinx + cosx) = sin2x - sinx
