Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

ĐKXĐ: \(sin2x\ne-\dfrac{1}{2}\)

Ta có: \(sin3x+cos3x=3sinx-4sin^3x+4cos^3x-3cosx\)

\(=\left(cosx-sinx\right)\left(4+2sin2x\right)-3\left(cosx-sinx\right)\)

\(=\left(cosx-sinx\right)\left(1+2sin2x\right)\)

Nên pt trở thành:

\(5\left(sinx+\dfrac{\left(cosx-sinx\right)\left(1+2sin2x\right)}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5cosx=cos2x+3\)

\(\Leftrightarrow2cos^2x-5cosx+2=0\Rightarrow\left[{}\begin{matrix}cosx=2\left(loại\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

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

- Đáp án D

- Đáp án C nha

\(cos5x+cos2x+2sin3x.sin2x=0\)

\(\Leftrightarrow cos5x+cos2x+cosx-cos5x=0\)

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

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

\(\Rightarrow x=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\) \(\Rightarrow\dfrac{\pi}{3}+\pi+\dfrac{5\pi}{3}=3\pi\)

ĐKXĐ: \(cos2x\ne0\)

\(\dfrac{cos4x}{cos2x}=\dfrac{sin2x}{cos2x}\)

\(\Rightarrow cos4x=sin2x\)

\(\Leftrightarrow cos4x=cos\left(\dfrac{\pi}{2}-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\Rightarrow x=\left\{\dfrac{\pi}{12};\dfrac{5\pi}{12}\right\}\) có 2 nghiệm

\(2cos5x.cosx=2cos4x.cos2x+6cos^2x+2\)

\(\Leftrightarrow cos6x+cos4x=cos6x+cos2x+3\left(1+cos2x\right)+2\)

\(\Leftrightarrow cos4x=4cos2x+5\)

\(\Leftrightarrow2cos^22x-1=4cos2x+5\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow2x=\pi+k2\pi\)

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

\(\Rightarrow x=-\dfrac{\pi}{2};\dfrac{\pi}{2}\)

- Đáp án B

\(\dfrac{sin3x+cos3x}{1+2sin2x}=\dfrac{3sinx-4sin^3x+4cos^3x-3cosx}{1+2sin2x}\)

\(=\dfrac{\left(cosx-sinx\right)\left(4+2sin2x\right)-3\left(cosx-sinx\right)}{1+2sin2x}=\dfrac{\left(cosx-sinx\right)\left(1+2sin2x\right)}{1+2sin2x}=cosx-sinx\)

Pt trở thành:

\(sinx+cosx-sinx=\dfrac{3+cos2x}{5}\)

\(\Leftrightarrow5cosx=3+2cos^2x-1\)

\(\Leftrightarrow2cos^2x-5cosx+2=0\Rightarrow\left[{}\begin{matrix}cosx=2\left(loại\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow x=\dfrac{\pi}{3};\dfrac{5\pi}{3}\)

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

\(\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sin2x}=2\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\dfrac{\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)}{sin2x}=2\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\\sinx-\sqrt{3}cosx=2sin2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\\\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\sin\left(x-\dfrac{\pi}{3}\right)=sin2x\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (đều thỏa mãn ĐKXĐ)