Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

VD15:

Pt \(\Leftrightarrow2\left(1-sin^2x\right)=sinx+1\)

\(\Leftrightarrow-2sin^2x-sinx+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{2}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

CD16:

Pt \(\Leftrightarrow\dfrac{1-cos6x}{2}-\dfrac{1-cos4x}{2}-\dfrac{1-cos2x}{2}=0\)

\(\Leftrightarrow-1-cos6x+cos4x+cos2x=0\)

\(\Leftrightarrow-\left(1+cos6x\right)+\left(cos4x+cos2x\right)=0\)

\(\Leftrightarrow-2.cos^23x+2.cos3x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cos3x=cosx\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

VD17:

Pt\(\Leftrightarrow2.sin3x.cosx=2.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

Pt \(\Leftrightarrow\left(2sin^2x-2\sqrt{2}sinx+1\right)+\left(3tan^2x-2\sqrt{3}tanx+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tanx-1\right)^2=0\)

Thấy \(VT\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tanx=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\) (Biểu diễn các cung này trên đường tròn lượng giác thấy không có cung nào trùng nhau) \(\Rightarrow\) Không tồn tại x để dấu bằng xảy ra

Vậy pt vô nghiệm

Pt\(\Leftrightarrow\dfrac{1}{2}\left(cos2004x+cos2006x\right)=1\)

\(\Leftrightarrow cos2004x+cos2006x=2\)

Có \(\left\{{}\begin{matrix}cos2004x\le1\\cos2006x\le1\end{matrix}\right.\)\(\Rightarrow cos2004x+cos2006x\le2\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}cos2004x=1\\cos2006x=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\pi}{1002}\\x=\dfrac{k\pi}{1003}\end{matrix}\right.\)\(\Rightarrow\dfrac{k\pi}{1002}=\dfrac{k\pi}{1003}\Leftrightarrow k=0\Rightarrow x=0\)

Vậy x=0 (Người chơi hệ đêm khuya à)

\(\Leftrightarrow2^{cosx}+cosx=2^{sinx}+sinx\) (1)

Xét hàm \(f\left(t\right)=2^t+t\)

\(f\left(t\right)\) đồng biến trên R \(\Rightarrow f\left(t_1\right)=f\left(t_2\right)\) khi và chỉ khi \(t_1=t_2\)

Nên (1) tương đương:

\(cosx=sinx\Leftrightarrow tanx=1\)

\(\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

Đk:\(x\ne\dfrac{k\pi}{2}\)

Pt\(\Leftrightarrow sin2x-cos2x=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\)

\(\Leftrightarrow sin2x-cos2x=\dfrac{1}{cosx.sinx}\)

\(\Leftrightarrow sin2x-cos2x=\dfrac{1}{\dfrac{1}{2}.sin2x}\)

\(\Leftrightarrow sin^22x-cos2x.sin2x=2\)

\(\Leftrightarrow sin^22x-cos2x.sin2x=2\left(sin^22x+cos^22x\right)\)

\(\Leftrightarrow-sin^22x-cos2x.sin2x-2cos^2x=0\) (vô nghiệm)

Vậy pt vô nghiệm

a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)

\(\Leftrightarrow3cos4x-cos6x-2=0\)

Đặt \(t=2x\)

Pttt:\(3cos2t-cos3t-2=0\)

\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)

\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))

\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)

\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)

\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))

Vậy...

a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)

\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)

\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)

\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))

\(\Rightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)

Vậy...

d)

PT \(\Leftrightarrow sin^2x-3sinx.cosx=sin^2x+cos^2x\)

\(\Leftrightarrow cos^2x+3sinx.cosx=0\)

\(\Leftrightarrow cosx\left(cosx+3sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-3sinx\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\tanx=-\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=arc.tan\left(-\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

j)\(\left(sinx-sin^2x\right)\left(sinx+2cosx\right)=\sqrt{3}\left(1+sinx\right)\left(1-sinx\right)^2\)

\(\Leftrightarrow sinx\left(1-sinx\right)\left(sinx+2cosx\right)=\sqrt{3}\left(1-sin^2x\right)\left(1-sinx\right)\)

\(\Leftrightarrow\left(1-sinx\right)\left[sin^2x+2sinx.cosx-\sqrt{3}\left(1-sin^2x\right)\right]=0\)

\(\Leftrightarrow\left(1-sinx\right)\left[sin^2x+2sinx.cosx-\sqrt{3}cos^2x\right]=0\)

\(\Leftrightarrow\left(1-sinx\right)\left[sinx-\left(-1-\sqrt{1+\sqrt{3}}\right)cosx\right]\left[sinx-\left(-1+\sqrt{1+\sqrt{3}}\right)cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\left(-1-\sqrt{1+\sqrt{3}}\right)cosx\\sinx=\left(-1+\sqrt{1+\sqrt{3}}\right)cosx\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\tan=-1-\sqrt{1+\sqrt{3}}\\tan=-1+\sqrt{1+\sqrt{3}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=arc.tan\left(-1-\sqrt{1+\sqrt{3}}\right)+k\pi\\x=arc.tan\left(-1+\sqrt{1+\sqrt{3}}\right)+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

Vậy...