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

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

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

1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)

=1

2: \(sin^4x-cos^4x\)

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

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

1.

\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)

2.

\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)

4.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

3. ĐKXĐ: ...

\(\Leftrightarrow\frac{sin\left(x-15\right)}{cos\left(x-15\right)}=\frac{3sin\left(x+15\right)}{cos\left(x+15\right)}\)

\(\Leftrightarrow sin\left(x-15\right)cos\left(x+15\right)=3sin\left(x+15\right)cos\left(x-15\right)\)

\(\Leftrightarrow sin2x-sin30^0=3\left[sin2x+sin30^0\right]\)

\(\Leftrightarrow sin2x-\frac{1}{2}=3sin2x+\frac{3}{2}\)

\(\Leftrightarrow sin2x=-1\)

\(\Leftrightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)

5.

\(sin6x+sin2x+sin4x=0\)

\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)

\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)

6. ĐKXĐ; ...

\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)

\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)

\(\Leftrightarrow tan3x=1\)

\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Câu 1 đề vẫn có vấn đề:

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)

\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)

Có thể coi như ko thể rút gọn tiếp

2.

\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

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

\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

1.

\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))

2.

Đề không đúng

3.

ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan2x=tanx\)

\(\Rightarrow2x=x+k\pi\)

\(\Rightarrow x=k\pi\)

4.

\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)

\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))

5.

\(cos3x=sin5x\)

\(\Leftrightarrow sin5x=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) (\(k\in Z\))

a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)

=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi

=>x=pi/8+kpi hoặc x=-pi/8+kpi

b: \(\Leftrightarrow sinx=sin\left(\dfrac{pi}{2}-3x\right)\)

=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi

=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi

=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi

d: \(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=-sin\left(3x+\dfrac{pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=sin\left(-3x-\dfrac{pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=cos\left(3x+\dfrac{3}{4}pi\right)\)

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

=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi

=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2

e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=0\)

=>x-pi/3=kpi

=>x=kpi+pi/3

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)

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

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

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)