Bài 4: Ôn tập chương Hàm số lượng giác và phương trình lượng giác

\(\Leftrightarrow1-2sin^2x+\sqrt{3}cosx+5sinx=2\sqrt{3}sinx.cosx+3\)

\(\Leftrightarrow2sin^2x-5sinx+2+\sqrt{3}cosx\left(2sinx-1\right)=0\)

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

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

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

\(\Leftrightarrow\left(2sinx-1\right)\left[sin\left(x+\dfrac{\pi}{3}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sin\left(x+\dfrac{\pi}{3}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(tanx.cosx+tanx\right)}{tanx-tanx.cosx}-2cosx=2\)

\(\Leftrightarrow\dfrac{3tanx\left(cosx+1\right)}{tanx\left(1-cosx\right)}-2cosx=2\)

\(\Leftrightarrow\dfrac{3cosx+3}{1-cosx}-2cosx=2\)

\(\Rightarrow3cosx+3-2cosx\left(1-cosx\right)=2\left(1-cosx\right)\)

\(\Leftrightarrow2cos^2x+3cosx+1=0\)

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

\(\Rightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(sin2x.cosx+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow sinx\left(2cos^2x-1\right)+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow sinx.cos2x+cos2x.cosx+2cos2x=0\)

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

\(\Leftrightarrow cos2x\left(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(\dfrac{1-cos2x}{2}\right)^2+\left[\dfrac{1+cos\left(2x+\dfrac{\pi}{2}\right)}{2}\right]^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left(1-cos2x\right)^2+\left(1-sin2x\right)^2=1\)

\(\Leftrightarrow cos^22x-2cos2x+1+sin^22x-2sin2x+1=1\)

\(\Leftrightarrow cos2x+sin2x=1\)

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

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

\(\Leftrightarrow...\)

\(\Leftrightarrow cos3x+sin7x=1-cos\left(\dfrac{\pi}{2}+5x\right)-1-cos9x\)

\(\Leftrightarrow cos3x+sin7x=sin5x-cos9x\)

\(\Leftrightarrow cos9x+cos3x+sin7x-sin5x=0\)

\(\Leftrightarrow2cos6x.cos3x+2cos6x.sinx=0\)

\(\Leftrightarrow cos6x\left(cos3x+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x=-sinx=cos\left(\dfrac{\pi}{2}+x\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}-\dfrac{1+cos16x}{2}=cos10x\)

\(\Leftrightarrow cos16x+cos4x+2cos10x=0\)

\(\Leftrightarrow2cos10x.cos6x+2cos10x=0\)

\(\Leftrightarrow cos10x\left(cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

Đặt \(\dfrac{x}{2}-\dfrac{\pi}{4}=t\Rightarrow\dfrac{x}{2}=t+\dfrac{\pi}{4}\)

\(\Rightarrow\dfrac{5x}{2}-\dfrac{\pi}{4}=5\left(t+\dfrac{\pi}{4}\right)-\dfrac{\pi}{4}=5t+\pi\)

\(\dfrac{3x}{2}=3\left(t+\dfrac{\pi}{4}\right)=3t+\dfrac{3\pi}{4}\)

Pr trở thành:

\(sin\left(5t+\pi\right)-cost=\sqrt{2}cos\left(3t+\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow-sin5t-cost=\sqrt{2}cos3t.cos\left(\dfrac{3\pi}{4}\right)-\sqrt{2}.sin3t.sin\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow-sin5t-cost=-cos3t-sin3t\)

\(\Leftrightarrow sin5t-sin3t+cost-cos3t=0\)

\(\Leftrightarrow2cos4t.sint+2sin2t.sint=0\)

\(\Leftrightarrow sint\left(cos4t+sin2t\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sint=0\\cos4t=-sin2t=cos\left(\dfrac{\pi}{2}+2t\right)\end{matrix}\right.\)

...

`sin^3 x cos x-sin x cos^3 x=\sqrt{2}/8`

`<=>sin x cos x(sin^2 x-cos^2 x)=\sqrt{2}/8`

`<=>-sin 2x cos 2x=\sqrt{2}/4`

`<=>sin 4x=-\sqrt{2}/2`

`<=>[(4x=-\pi/4+k2\pi),(4x=[5\pi]/4+k2\pi):}`

`<=>[(x=-\pi/16+k\pi/2),(x=[5\pi]/16+k\pi/2):}`   `(k in ZZ)`

a.

Đặt \(x+\dfrac{\pi}{3}=t\Rightarrow x=t-\dfrac{\pi}{3}\)

\(\Rightarrow\dfrac{\pi}{6}-x=\dfrac{\pi}{6}-\left(t-\dfrac{\pi}{3}\right)=\dfrac{\pi}{2}-t\)

Pt trở thành:

\(cos2t+4cos\left(\dfrac{\pi}{2}-t\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2sin^2t+4sint=\dfrac{5}{2}\)

\(\Leftrightarrow4sin^2t-8sint+3=0\)

\(\Rightarrow\left[{}\begin{matrix}sint=\dfrac{3}{2}>1\left(loại\right)\\sint=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(tan^2x+cot^2x+2\left(tanx+cotx\right)=6\)

\(\Leftrightarrow\left(tanx+cotx\right)^2+2\left(tanx+cotx\right)-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx+cotx=2\\tanx+cotx=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx+\dfrac{1}{tanx}=2\\tanx+\dfrac{1}{tanx}=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tan^2x-2tanx+1=0\\tan^2x+4tanx+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{5\pi}{12}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(cos^2x+\dfrac{1}{cos^2x}=cosx+\dfrac{1}{cosx}\)

\(\Leftrightarrow\left(cosx+\dfrac{1}{cosx}\right)^2-\left(cosx+\dfrac{1}{cosx}\right)-2=0\)

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

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

\(\Rightarrow cosx=1\)

\(\Rightarrow x=k2\pi\)