Bài 3: Một số phương trình lượng giác thường gặp

Sửa đề: \(sin7x-cos2x=\sqrt{3}\left(sin2x-cos7x\right)\)

\(\Leftrightarrow sin7x\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot cos2x=\dfrac{\sqrt{3}}{2}\cdot sin2x-\dfrac{\sqrt{3}}{2}\cdot cos7x\)

\(\Leftrightarrow sin\left(7x+\dfrac{pi}{3}\right)=sin\left(2x+\dfrac{pi}{6}\right)\)

=>7x+pi/3=2x+pi/6+k2pi hoặc 7x+pi/3=-2x+5/6pi+k2pi

=>5x=-pi/6+k2pi hoặc 9x=1/2pi+k2pi

=>x=-pi/30+k2pi/5 hoặc x=1/18pi+k2pi/9

1.

\(2sin\left(-x\right)+4-6sin\left(-x\right)=0\)

\(\Leftrightarrow-2sinx+4+6sinx=0\)

\(\Leftrightarrow4sinx-4=0\)

\(\Leftrightarrow sinx=1\)

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

2.

ĐKXĐ; \(x\ne\left\{\dfrac{\pi}{4}+\dfrac{k\pi}{2};k\pi\right\}\)

\(\left(tan2x+1\right)\left(2cotx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tan2x=-1\\cotx=\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+k\pi\\x=arcot\left(\dfrac{3}{2}\right)+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=arccot\left(\dfrac{3}{2}\right)+k\pi\end{matrix}\right.\)

17.

\(cosx.cos3x=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)

\(\Leftrightarrow2cos^22x-1+cos2x=-1\)

\(\Leftrightarrow2cos^22x-cos2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

17: \(\Leftrightarrow\dfrac{1}{2}\cdot\left(cos\left(-2x\right)+cos4x\right)=\dfrac{-1}{2}\)

=>\(cos2x+cos4x=-1\)

=>\(2cos^22x-1+cos2x=-1\)

=>cos2x(2 cos 2x+1)=0

=>cos2x=0 hoặc cos2x=-1/2

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{pi}{2}+kpi\\2x=\dfrac{2}{3}pi+k2pi\\2x=-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{4}+\dfrac{kpi}{2}\\x=\dfrac{1}{3}pi+kpi\\x=-\dfrac{1}{3}pi+kpi\end{matrix}\right.\)

19: \(\Leftrightarrow4-4cos^2x+\left(2\sqrt{3}+2\right)cosx-4-\sqrt{3}=0\)

\(\Leftrightarrow4cos^2x-\left(2\sqrt{3}+2\right)cosx+\sqrt{3}=0\)

\(\text{Δ}=16-8\sqrt{3}\)

Pt sẽ có hai nghiệm là:

\(\left\{{}\begin{matrix}cosx=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{8}=\dfrac{4}{8}=\dfrac{1}{2}\\cosx=\dfrac{2\sqrt{3}+2+2\sqrt{3}-2}{8}=\dfrac{4\sqrt{3}}{8}=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{pi}{6}+k2pi\\x=\pm\dfrac{pi}{3}+k2pi\end{matrix}\right.\)

18.

\(3+sinx.sin3x=3cos2x\)

\(\Leftrightarrow3+sinx.sin3x=3\left(1-2sin^2x\right)\)

\(\Leftrightarrow6sin^2x+sinx.sin3x=0\)

\(\Leftrightarrow6.sin^2x+sinx\left(3sinx-4sin^3x\right)=0\)

\(\Leftrightarrow6sin^2x+sin^2x\left(3-4sin^2x\right)=0\)

\(\Leftrightarrow sin^2x\left(9-4sin^2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\sin^2x=\dfrac{9}{4}\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=k\pi\)

Em nhập lại nội dung câu hỏi nhé!

??

1: =>\(cos3x=-\dfrac{\sqrt{2}}{2}\)

=>3x=3/4pi+k2pi hoặc 3x=-3/4pi+k2pi

=>x=1/4pi+k2pi/3 hoặc x=-1/4pi+k2pi/3

5: =>2 cot3x=-5

=>cot3x=-5/2

=>3x=arccot(-5/2)+kpi

=>x=1/3(arccot(-5/2)+kpi)

1: =>cos3x=-căn 2(vô lý)

2: \(\Leftrightarrow sin\left(x+\dfrac{pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)

=>x+pi/4=pi/3+k2pi hoặc x+pi/4=2/3pi+k2pi

=>x=pi/12+k2pi hoặc x=5/12pi+k2pi

3: \(\Leftrightarrow tan\left(x+15^0\right)=\dfrac{1}{\sqrt{3}}\)

=>x+15 độ=30 độ+k*180 độ

=>x=15 độ+k*180 độ

6: =>2 cot 3x=-5

=>cot 3x=-5/2

=>3x=arccot(-5/2)+kpi

=>x=1/3(arccot(-5/2)+kpi)

cot(x-pi/6)=căn 3

=>x-pi/6=pi/6+kpi

=>x=1/3pi+kpi

Khi k=0 thì x=1/3pi

=>Chọn B

Ta có:`cos x/3 = 0`

`\iff x/3 = \pi/2 + k\pi `

`\iff x = \frac{3\pi}{2} + k3\pi ( k \in ZZ)`

Do đó A là đáp án đúng.

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

`<=>1/2sin 3x-\sqrt{3}/2 cos 3x=sin 2x`

`<=>sin (3x-\pi/3)=sin 2x`

`<=>[(3x-\pi/3=2x+k2\pi),(3x-\pi/3=\pi-2x+k2\pi):}`

`<=>[(x=\pi/3+k2\pi),(x=[4\pi]/15+k[2\pi]/5):}`    `(k in ZZ)`