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

\(\Leftrightarrow cos2x+2cosx+cos^2x+2sin^2x=0\\ \Leftrightarrow cos2x+2cosx+cos^2x+1-cos2x=0\\ \Leftrightarrow\left(cos^2x+1\right)=0\Leftrightarrow x=\pi+k2\pi\)

\(3tan\cdot\left(4x-\dfrac{\pi}{3}\right)+\sqrt{3}=0\)

\(\Rightarrow tan\left(4x-\dfrac{\pi}{3}\right)=-\dfrac{1}{\sqrt{3}}=tan\left(-\dfrac{\pi}{6}\right)\)

\(\Rightarrow4x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k\pi\)\(\Rightarrow4x=\dfrac{\pi}{6}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{24}+k\dfrac{\pi}{4}\left(k\in Z\right)\)

ĐK: \(x\ne\dfrac{\pi}{6}+k2\pi;x\ne\dfrac{5\pi}{6}+k2\pi\)

\(\dfrac{cosx-\sqrt{3}sinx}{sinx-\dfrac{1}{2}}=0\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=0\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Đối chiếu điều kiện ta được \(x=-\dfrac{5\pi}{6}+k2\pi\).

a) \(2sin2x-1=0\Rightarrow sin2x=\dfrac{1}{2}=sin\dfrac{\pi}{6}\)

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

2)  \(2sin3x+\sqrt{3}=0\Rightarrow sin3x=-\dfrac{\sqrt{3}}{2}=sin\left(-\dfrac{\pi}{3}\right)\)

    \(\Rightarrow\left[{}\begin{matrix}3x=-\dfrac{\pi}{3}+k2\pi\\3x=\pi-\left(-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)  Bạn tự tìm x nhé.

c) \(2cos2x-3cosx+1=0\Rightarrow2(2cos^2x-1)-3cosx+1=0\)

   \(\Rightarrow4cos^2x-3cosx-1=0\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{4}\end{matrix}\right.\)

Bạn tự tìm x nhé.

f) \(cos2x+3sinx-2=0\Rightarrow1-2sin^2x+3sinx-2=0\)

   \(-2sin^2x+3sinx-1=0\) \(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

Bạn tự tìm x nhé.

a) \(2sim\left(2x-\dfrac{\pi}{3}\right)-1=0\Rightarrow sin\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}=sin\dfrac{\pi}{6}\)

    \(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.

d) \(\sqrt{3}sinx-cosx=\sqrt{2}\) \(\Rightarrow2sin\left(x-\dfrac{\pi}{6}\right)=\sqrt{2}\)

     \(\Rightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}=sin\dfrac{\pi}{4}\)

     \(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.