\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)

\(=1+\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=1+\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}\)

\(=1+\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=1+\frac{2sin2x}{sin2x}=1+2=3\)

\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)

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

\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)

\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)

\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)

\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(A=\frac{tana+tanb}{tan\left(a+b\right)}-\frac{tana-tanb}{tan\left(a-b\right)}=\frac{tana+tanb}{\frac{tana+tanb}{1-tana\cdot tanb}}-\frac{tana-tanb}{\frac{tana-tanb}{1+tana\cdot tanb}}\\ \Leftrightarrow A=1-tana\cdot tanb-1-tana\cdot tanb=-2tana\cdot tanb\)

\(B=\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}\\ B=\frac{cos^3x-4cos^3x+3cosx}{cosx}+\frac{sin^3x+3sinx-4sin^3x}{sinx}\\ B=\frac{-3cos^3x+3cosx}{cosx}+\frac{-3sin^3x+3sinx}{sinx}\\ B=\frac{cosx\left(-3cos^2x+3\right)}{cosx}+\frac{sinx\left(-3sin^2x+3\right)}{sinx}\\ B=-3cos^2x+3-3sin^2x+3=6-3\left(sin^2x+cos^2x\right)=6-3=3\)

~~~~~~~~chúc bạn học tốt~~~~~~~~~~

\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)

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

\(\frac{cosx+cos3x+cos2x+cos4x}{sinx+sin3x+sin2x+sin4x}=\frac{2cosx.cos2x+2cosx.cos3x}{2sin2x.cosx+2sin3x.cosx}=\frac{2cosx\left(cos2x+cos3x\right)}{2cosx\left(sin2x+sin3x\right)}\)

\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)

a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)

b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx

⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x

⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x

⇔ 4sin2x + (sinx + cosx) . sin2x = 0

⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)

⇔ sin2x = 0

c, 2cos³x = sin3x

⇔ 2cos³x = 3sinx - 4sin³x

⇔ 4sin³x + 2cos³x - 3sinx(sin²x + cos²x) = 0

⇔ sin³x + 2cos³x - 3sinx.cos²x = 0

Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình

Xét cosx ≠ 0 chia cả 2 vế cho cos³x ta được : 

tan³x + 2 - 3tanx = 0

⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)

d, cos²x - \(\sqrt{3}sin2x\) = 1 + sin²x

⇔ cos²x - sin²x - \(\sqrt{3}sin2x\) = 1

⇔ cos2x - \(\sqrt{3}sin2x\) = 1

⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)

⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)

e, cos³x + sin³x = 2cos⁵x + 2sin⁵x

⇔ cos³x (1 - 2cos²x) + sin³x (1 - 2sin²x) = 0

⇔ cos³x . (- cos2x) + sin³x . cos2x = 0

⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)

7.

ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow cos2x\ne0\)

Phương trình tương đương:

\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)

\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)

\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)

\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)

\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)

\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)

\(\Leftrightarrow2cos^44x-cos^24x-1=0\)

\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)

\(\Leftrightarrow cos^24x-1=0\)

\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)

\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

1.

\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)

\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)

\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(1-t^2-4t+4=0\)

\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)

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

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

2.

\(\Leftrightarrow\left(sinx-1\right)^2+1=sin^23x\)

Ta có \(VT\ge1\) trong khi \(VP\le1\) với mọi x

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sinx-1=0\\sin^23x=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)

3.

\(\Leftrightarrow-2cos2x.sinx-2sin2x=2\sqrt{2}\)

\(\Leftrightarrow cos2x.sinx+sin2x=-\sqrt{2}\)

Ta có:

\(VT^2=\left(cos2x.sinx+sin2x.1\right)^2\le\left(cos^22x+sin^22x\right)\left(sin^2x+1\right)\le1\left(1+1\right)=2\)

\(\Rightarrow VT\ge-\sqrt{2}\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sinx=1\\cos2x=sinx.sin2x\end{matrix}\right.\) (ko tồn tại x thỏa mãn)

Vậy pt vô nghiệm