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

Hàm số xác định khi:

\(\left\{{}\begin{matrix}\dfrac{1-sinx}{2+cosx}\ge0\\2+cosx\ne0\\1-cosx>0\end{matrix}\right.\Leftrightarrow1-cosx\ne0\Leftrightarrow x\ne k2\pi\)

sin2x + 1 - 2sin²x + sinx + cosx = 0

⇔ sin2x + cos2x + sinx + cosx = 0

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

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

⇔ \(2sin\left(\dfrac{3x}{2}+\dfrac{\pi}{4}\right).cos\dfrac{x}{2}=0\)

⇔ \(\left[{}\begin{matrix}sin\left(3x+\dfrac{\pi}{4}\right)=0\\cos\dfrac{x}{2}=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k.\dfrac{\pi}{3}\\x=\pi+k.2\pi\end{matrix}\right.\) , k ∈ Z

\(2sin2x+sinx.cosx-cos^2x+1=0\)

\(\Leftrightarrow4sin2x+2sinx.cosx-2cos^2x+2=0\)

\(\Leftrightarrow4sin2x+sin2x-cos2x=-1\)

\(\Leftrightarrow5sin2x-cos2x=-1\)

\(\Leftrightarrow\sqrt{26}\left(\dfrac{5}{\sqrt{26}}sin2x-\dfrac{1}{\sqrt{26}}cos2x\right)=-1\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{1}{\sqrt{26}}\right)=\dfrac{1}{\sqrt{26}}\)

\(\Leftrightarrow2x+arccos\dfrac{1}{\sqrt{26}}=\pm arccos\dfrac{1}{\sqrt{26}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-arccos\dfrac{1}{\sqrt{26}}+k\pi\end{matrix}\right.\)

ĐKXĐ: \(x\ne k\pi\)

\(sin7x=sin^2x+2sinx.cos2x+2sinx.cos4x+2sinx.cos6x\)

\(\Leftrightarrow sin7x=sin^2x+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(\Leftrightarrow sin7x=sin^2x-sinx+sin7x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\sinx=1\end{matrix}\right.\)

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

\(cos^2x-sin^2x=sinx-cosx\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)=-\left(cosx-sinx\right)\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

C

C

C

1.

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sinx=-1\)

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

2.

\(sin2x\left(2sinx-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\2sinx-\sqrt{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(sin\left(\dfrac{x+\pi}{5}\right)=-\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x+\pi}{5}=-\dfrac{\pi}{6}+k2\pi\\\dfrac{x+\pi}{5}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11\pi}{6}+k10\pi\\x=\dfrac{29\pi}{6}+k10\pi\end{matrix}\right.\)

4.

\(2sin\left(2x-10^0\right)=\sqrt{3}\Rightarrow sin\left(2x-10^0\right)=\dfrac{\sqrt{3}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+k180^0\end{matrix}\right.\)

\(\Rightarrow x=\left\{-145^0;35^0;-115^0;65^0\right\}\) có 4 nghiệm

C3 

\(sin\left(\dfrac{x+\Pi}{5}\right)=sin\left(\dfrac{-\pi}{6}\right)\)

<=>\(^{\left[{}\begin{matrix}\dfrac{x+pi}{5}=\dfrac{-pi}{6}+k2pi\\\dfrac{x+pi}{5}=\dfrac{7pi}{6}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{5}=-\dfrac{11pi}{30}+k2pi\\\dfrac{x}{5}=\dfrac{29pi}{30}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11pi}{6}+\dfrac{k2pi}{5}\\x=\dfrac{29pi}{6}+\dfrac{k2pi}{5}\end{matrix}\right.\)

Câu 3: B

Câu 4: B