\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)

\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)

\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)

\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)

\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)

\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)

b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)

c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)

\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)

d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)

\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)

e/

@Nguyễn Việt Lâm giúp em với ạ

a/ \(4cos^3x-3cosx-4\left(2cos^2x-1\right)+3cosx-4=0\)

\(\Leftrightarrow4cos^3x-8cos^2x=0\)

\(\Leftrightarrow4cos^2x\left(cosx-2\right)=0\)

\(\Leftrightarrow cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(0< \frac{\pi}{2}+k\pi< 14\Rightarrow-\frac{1}{2}< k< \frac{14-\frac{\pi}{2}}{\pi}\Rightarrow k=\left\{0;1;2;3\right\}\)

\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{7\pi}{2}\right\}\)

b/ Bạn coi lại đề, cái ngoặc thứ 2 thiếu \(\left(2cos\left(???\right)+cosx\right)\)

c/ Bạn coi lại đề, có 2 số hạng \(cos2x\) xuất hiện ở vế trái, cấp 3 chắc ko ai cho kiểu vậy đâu, nếu đúng thế thì người ta cộng luôn thành \(2cos2x\) cho rồi

a, \(sin^2x-4sinx+3=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

b, \(2cos^2-cosx-1=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

c, \(3sin^2x-2cosx+2=0\)

\(\Leftrightarrow3-3sin^2x+2cosx-5=0\)

\(\Leftrightarrow3cos^2x+2cosx-5=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(3cosx+5\right)=0\)

\(\Leftrightarrow cosx=1\)

\(\Leftrightarrow x=k2\pi\)

\(VT=\dfrac{2\cdot sin2x+2\cdot sin2x\cdot cos2x}{2\cdot\left(cosx+cos3x\right)}\)

\(=\dfrac{2\cdot sin2x\left(1+cos2x\right)}{2\cdot\left(cosx+cos3x\right)}\)

\(=\dfrac{sin2x\cdot\left(1+2cos^2x-1\right)}{cosx+cos3x}\)

\(=\dfrac{sin2x\cdot2\cdot cos^2x}{2\cdot cos\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)}\)

\(=\dfrac{sin2x\cdot cos^2x}{cosx\cdot cos2x}=\dfrac{sin2x}{cos2x}\cdot cosx=tan2x\cdot cosx\)

Lời giải:

Áp dụng các công thức lượng giác:

\(1+\cos x+\cos 2x+\cos 3x\)

\(=(1+\cos 2x)+(\cos x+\cos 3x)\)

\(=2\cos ^2x+2\cos 2x\cos x\)

\(=2\cos x(\cos x+\cos 2x)=2\cos x(\cos x+\cos ^2x-\sin ^2x)\)

\(=2\cos x(\cos x+2\cos ^2x-1)\)

\(\Rightarrow \frac{1+\cos x+\cos 2x+\cos 3x}{2\cos ^2x+\cos x-1}=\frac{2\cos x(\cos x+2\cos ^2x-1)}{2\cos ^2x+\cos x-1}=2\cos x\)

Vậy \(2\cos x=\frac{2}{3}(3-\sqrt{3})\sin x\)

\(\Leftrightarrow \sqrt{3}\cos x=(\sqrt{3}-1)\sin x\)

\(\Rightarrow \tan x=\frac{\sin x}{\cos x}=\frac{\sqrt{3}}{\sqrt{3}-1}\Rightarrow x=k\pi +\arctan \frac{\sqrt{3}}{\sqrt{3}-1}\)