Bài 2: Phương trình lượng giác cơ bản

a.

\(\Leftrightarrow2-2cos\left(4x-2\right)=1\)

\(\Leftrightarrow cos\left(4x-2\right)=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}4x-2=\dfrac{\pi}{3}+k2\pi\\4x-2=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{1}{2}-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{2}cos6x+\dfrac{1}{2}-\dfrac{1}{2}cos8x=1\)

\(\Leftrightarrow cos6x=cos8x\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=6x+k2\pi\\8x=-6x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{7}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{k\pi}{7}\)

`a)4cos^2(2x-1)=1`

`<=>4[1+cos(4x-2)]/2=1`

`<=>2(1+cos(4x-2))=1`

`<=>2cos(4x-2)=-1`

`<=>cos(4x-2)=-1/2`

`<=>[(4x-2=[2\pi]/3+k2\pi),(4x-2=[-2\pi]/3+k2\pi):}`

`<=>[(x=1/2+\pi/6+k\pi/2),(x=1/2-\pi/6+k\pi/2):}`   `(k in ZZ)`

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`c)cos^2 3x+sin^2 4x=1`

`<=>[1+cos 6x]/2+[1-cos 8x]/2=1`

`<=>1+cos 6x+1-cos 8x=2`

`<=>cos 8x=cos 6x`

`<=>[(8x=6x+k2\pi),(8x=-6x+k2\pi):}`

`<=>[(x=k\pi),(x=k\pi/7):}`    `(k in ZZ)`

1: =>cosx=căn 3/2

=>x=pi/3+k2pi hoặc x=-pi/3+k2pi

2: =>cos2x=1/3

=>2x=arccos(1/3)+k2pi hoặc 2x=-arccos(1/3)+k2pi

=>x=1/2arcos(1/3)+kpi hoặc x=-1/2arccos(1/3)+kpi

3: =>cos(x-pi/3)=-căn 2/2=3/4pi

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

=>x=13/12pi+k2pi hoặc x=-5/12pi+k2pi

4: =>cos(2/5pi-x/2)=-căn 3/2

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

=>x/2=-13/30pi-k2pi hoặc x/2=37/30pi-k2pi

=>x=-13/60pi-4kpi hoặc x=37/60pi-4kpi

\(\sin\left(x+\dfrac{\pi}{7}\right)+\sin\left(\dfrac{3\pi}{2}-\dfrac{\pi}{5}\right)=0\\ \Rightarrow\sin\left(x+\dfrac{\pi}{7}\right)+\sin\left(\dfrac{13\pi}{10}\right)=0\\ \Rightarrow\sin\left(x+\dfrac{\pi}{7}\right)=-\sin\left(\dfrac{13\pi}{10}\right)=sin\left(-\dfrac{13\pi}{10}\right)\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{7}=-\dfrac{13\pi}{10}+k2\pi\\x+\dfrac{\pi}{7}=\pi+\dfrac{13\pi}{10}+k2\pi\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{101\pi}{70}+k2\pi\\x=\dfrac{151\pi}{70}+k2\pi\end{matrix}\right.\)

Chọn B

\(2sinx+1=0\Leftrightarrow sinx=-\dfrac{1}{2}\)

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

ĐKXĐ: \(sin^2x+2\cdot sinx+4>=0\)

=>\(\left(sinx+1\right)^2+3>=0\)(luôn đúng)

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

`<=>sin` `\pi/3 cos x-cos` `\pi/3 sin x=3x`

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

`<=>` $\left[\begin{matrix} \dfrac{\pi}{3}-x=3x+k2\pi\\ \dfrac{\pi}{3}-x=\pi-3x+k2\pi\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\ x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.$   `(k in ZZ)`

Vậy ptr có `2` họ nghiệm `x=\pi/12-[k\pi]/2` ; `x=\pi/3+k\pi`  `(k in ZZ)`

\(\Leftrightarrow\cot^22x=\tan^2x\)

\(\Leftrightarrow\tan^2\left(\dfrac{\Pi}{2}-2x\right)=\tan^2x\)

TH1: \(\tan\left(\dfrac{\Pi}{2}-2x\right)=\tan\left(x\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< >\dfrac{\Pi}{2}+k\Pi\\\dfrac{\Pi}{2}-2x=x+k\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< >\dfrac{\Pi}{2}+k\Pi\\-3x=k\Pi-\dfrac{\Pi}{2}\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{3}\left(k\Pi-\dfrac{\Pi}{2}\right)\)

TH2: \(\tan\left(\dfrac{\Pi}{2}-2x\right)=\tan\left(-x\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x< >\dfrac{\Pi}{2}+k\Pi\\\dfrac{\Pi}{2}-2x=-x+k\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< >-\dfrac{\Pi}{2}-k\Pi\\-x=k\Pi-\dfrac{\Pi}{2}\end{matrix}\right.\Leftrightarrow x=-k\Pi+\dfrac{\Pi}{2}\)