Question

giải các pt

a) \(3cos4x-8cos^6x+2cos^2x+3=0\)

b) \(4+3sinx+sin^3x=3cos^2x+cos^6x\)

c) \(2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=cos2x-3\)

d) \(\frac{\sqrt{3}}{cos^2x}-tanx-2\sqrt{3}=sinx\left(1+tanx.tan\frac{x}{2}\right)\)

Answer 1

a/

\(\Leftrightarrow3\left(cos4x+1\right)+2cos^2x\left(1-4cos^4x\right)=0\)

\(\Leftrightarrow3\left(2cos^22x-1+1\right)+2cos^2x\left(1-2cos^2x\right)\left(1+2cos^2x\right)=0\)

\(\Leftrightarrow6cos^22x+\left(1+cos2x\right).\left(-cos2x\right)\left(2+cos2x\right)=0\)

Đặt \(cos2x=a\)

\(\Rightarrow6a^2-a\left(a+1\right)\left(a+2\right)=0\)

\(\Leftrightarrow a\left(-a^2+3a-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\\cos2x=2\left(l\right)\end{matrix}\right.\)

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

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

Answer 2

b/

\(\Leftrightarrow4+3sinx+sin^3x=3\left(1-sin^2x\right)+\left(1-sin^2x\right)^3\)

Đặt \(sinx=a\) ta được:

\(a^3+3a+4=3-3a^2+\left(1-a\right)^3\)

\(\Leftrightarrow a^3+3a^2+3a+1=\left(1-a\right)^3\)

\(\Leftrightarrow\left(a+1\right)^3=\left(1-a\right)^3\)

\(\Leftrightarrow a+1=1-a\)

\(\Leftrightarrow a=0\)

\(\Rightarrow sinx=0\Rightarrow x=k\pi\)

Answer 3

c/

ĐKXĐ: ...

\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)

\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)

\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)

\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)

\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pi+k2\pi\)

Answer 4

d/

ĐKXĐ: ...

Biến đôi biểu thức vế trái trước:

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

Do đó pt tương đương:

\(\sqrt{3}\left(1+tan^2x\right)-tanx-2\sqrt{3}=sinx.\frac{1}{cosx}\)

\(\Leftrightarrow\sqrt{3}tan^2x-2tanx-\sqrt{3}=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Sử dụng kết quả biến đổi trên làm câu c sẽ lẹ hơn cách cũ