b, \(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(=\dfrac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}\)

\(=\dfrac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx-cosx}{cosx}\right)^2}{\left(\dfrac{sinx+cosx}{cosx}\right)^2}\)

\(=\dfrac{\left(\dfrac{sinx}{cosx}-1\right)^2}{\left(\dfrac{sinx}{cosx}+1\right)^2}\)

\(=\dfrac{\left(tanx-tan\dfrac{\pi}{4}\right)^2}{\left(1+tanx.tan\dfrac{\pi}{4}\right)^2}\)

\(=tan^2\left(x-\dfrac{\pi}{4}\right)=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\)

1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)

\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)

\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )

b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)

\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)

\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)

\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )

c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)

\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)

\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)

\(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)

\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)

\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )

d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)

\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )

1/ \(3-4\sin^2=4\cos^2x-1\Leftrightarrow4\left(\sin^2x+\cos^2x\right)-4=0\Leftrightarrow4.1-4=0\left(ld\right)\Rightarrow dpcm\)

2/ \(\cos^4x-\sin^4x=\left(\cos^2x+\sin^2x\right)\left(\cos^2x-\sin^2x\right)=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x\)

3/ \(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x=1-2\sin^2x.\cos^2x\)

1.

\(\Leftrightarrow2cos2x+\sqrt{2}.\frac{\sqrt{2}}{2}=0\)

\(\Leftrightarrow cos2x=-\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\frac{\pi}{3};\frac{4\pi}{3};\frac{2\pi}{3};\frac{5\pi}{3}\right\}\)

2.

\(\Leftrightarrow sin4x-cos4x+sin4x+cos4x=\sqrt{6}\)

\(\Leftrightarrow2sin4x=\sqrt{6}\)

\(\Leftrightarrow sin4x=\frac{\sqrt{6}}{2}>1\)

Pt vô nghiệm

a: =>cos5x=cos(pi/2-3x)

=>5x=pi/2-3x+k2pi hoặc 5x=3x-pi/2+k2pi

=>8x=pi/2+k2pi hoặc 2x=-pi/2+k2pi

=>x=pi/16+kpi/8 hoặc x=-pi/4+kpi

b: sin4x=cos(x+pi/6)

=>sin4x=sin(pi/2-x-pi/6)

=>sin4x=sin(pi/3-x)

=>4x=pi/3-x+k2pi hoặc 4x=pi-pi/3+x+k2pi

=>5x=pi/3+k2pi hoặc 3x=2/3pi+k2pi

=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3

a) y=\(sin^4x+cos^4x-3=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-3=-2-\dfrac{1}{2}.sin^22x\)

Có \(0\le sin^22x\le1\)

\(\Leftrightarrow-2\ge y\ge-\dfrac{5}{2}\)

Min xảy ra \(\Leftrightarrow sin^22x=1\Leftrightarrow sin2x=1\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\left(k\in Z\right)\)

\(\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\left(k\in Z\right)\)

Max xảy ra \(\Leftrightarrow sin2x=0\Leftrightarrow2x=k\Pi\Leftrightarrow x=\dfrac{k\Pi}{2}\)

b, \(x\in\left[0;\pi\right]\)

=>\(sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\dfrac{\sqrt{2}}{2};1\right]\)

\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\sqrt{2};2\right]\)

\(\Rightarrow\left\{{}\begin{matrix}Miny=-\sqrt{2}\\Maxy=2\end{matrix}\right.\)

Min xảy ra \(\Leftrightarrow x=0\)

Max xảy ra \(\Leftrightarrow x=\dfrac{\pi}{2}\)