Sử dụng bộ gõ công thức ở chỗ khoanh đỏ trên khung soạn thảo cho người khác dễ đọc đi bạn:

1.

\(sin^2x+sin^23x=cos^2x+cos^23x\)

\(\Leftrightarrow sin^2x-cos^2x=cos^23x-sin^23x\)

\(\Leftrightarrow-cos2x=cos6x\)

\(\Leftrightarrow cos2x+cos6x=0\)

\(\Leftrightarrow2cos4x.cos2x=0\)

\(\Rightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

2.

\(cos\left(\frac{\pi}{3}+x\right)+cos\left(\frac{\pi}{3}-x\right)=1\)

\(\Leftrightarrow2cos\left(\frac{\pi}{3}\right).cosx=1\)

\(\Leftrightarrow cosx=1\)

\(\Rightarrow x=k2\pi\)

3.

\(sinx+cosx+\sqrt{2}sin2x=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)+\sqrt{2}sin2x=0\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-sin2x=sin\left(-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-2x+k2\pi\\x+\frac{\pi}{4}=\pi+2x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k2\pi}{3}\\x=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

4.

\(sin^4\frac{x}{2}+cos^4\frac{x}{2}=\frac{5}{8}\)

\(\Leftrightarrow\left(sin^2\frac{x}{2}+cos^2\frac{x}{2}\right)^2-2sin^2\frac{x}{2}.cos^2\frac{x}{2}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}\left(2sin\frac{x}{2}.cos\frac{x}{2}\right)^2=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^2x=\frac{5}{8}\)

\(\Leftrightarrow sin^2x=\frac{3}{4}\Rightarrow\left[{}\begin{matrix}sinx=\frac{\sqrt{3}}{2}\\sinx=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

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

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

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

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

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0

a: \(\Leftrightarrow2\cdot\sin3x\cdot\cos x-2\cos^2x=0\)

\(\Leftrightarrow\cos x\left(\sin3x-\cos x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\\sin3x=\cos x=\sin\left(\dfrac{\Pi}{2}-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\3x=\dfrac{\Pi}{2}-x+k2\Pi\\3x=\dfrac{\Pi}{2}+x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow\sin x+\sin5x+\sin^2x=0\)

\(\Leftrightarrow\sin x=0\)

hay \(x=k\Pi\)

1: =>sin^2(3x)=0

=>sin 3x=0

=>3x=kpi

=>x=kpi/3

2:

\(sinx=1-cos^2x=sin^2x\)

=>\(sin^2x-sinx=0\)

=>sin x(sin x-1)=0

=>sin x=0 hoặc sin x=1

=>x=pi/2+k2pi hoặc x=kpi

4:

sin 2x+sin x=0

=>sin 2x=-sin x=sin(-x)

=>2x=-x+k2pi hoặc 2x=pi+x+k2pi

=>x=pi+k2pi hoặc x=k2pi/3

5: =>cos(x+pi/3)=1/căn 2

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

=>x=-pi/12+k2pi hoặc x=-7/12pi+k2pi