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

\(\Leftrightarrow4sinx+2cosx+2=sinx-2cosx+3\)

\(\Leftrightarrow3sinx+4cosx=1\)

\(\Leftrightarrow\dfrac{3}{5}sinx+\dfrac{4}{5}cosx=\dfrac{1}{5}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{3}{5}=sin\varphi\\\dfrac{4}{5}=cos\varphi\end{matrix}\right.\)

\(pt\Leftrightarrow sin\varphi\cdot sinx+cos\varphi\cdot cos=\dfrac{1}{5}\)

\(\Leftrightarrow cos\cdot\left(\varphi-x\right)=\dfrac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\varphi-x=arc\cdot cos\dfrac{1}{5}+k2\pi\\\varphi-x=-arc\cdot cos\dfrac{1}{5}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\varphi+arc\cdot cos\dfrac{1}{5}+k2\pi\\x=\varphi-arc\cdot cos\dfrac{1}{5}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

1.

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

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(\Leftrightarrow2cosx-sinx-4sin^2x.cosx+2sin^3x=sin^3x+cos^3x\)

\(\Leftrightarrow sin^3x-cos^3x-4sin^2x.cosx+2cosx-sinx=0\)

- Với \(\left\{{}\begin{matrix}cosx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\) là nghiệm của pt

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(tan^3x-1-4tan^2x+2\left(1+tan^2x\right)-tanx\left(1+tan^2x\right)=0\)

\(\Leftrightarrow-2tan^2x-tanx+3=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{3}{2}\end{matrix}\right.\)

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

2.1

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

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

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

c.

\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)

Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm

d.

\(cosx-sinx=1\)

\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)

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

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

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

1.

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(m+1\right)^2+\left(-3\right)^2\ge m^2\)

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow3\left(\frac{1}{2}-\frac{1}{2}cos2x\right)+4m.sin2x-4=0\)

\(\Leftrightarrow8m.sin2x-3cos2x=5\)

Pt vô nghiệm khi: \(\left(8m\right)^2+\left(-3\right)^2< 5^2\)

\(\Leftrightarrow...\)

ĐKXĐ: \(sinx\ne0\)

\(2cos^2x-3cosx+1=sinx-2sinx^2cosx+2cos^2x.sinx\)

\(\Leftrightarrow2cos^2x\left(1-sinx\right)+1-sinx-3cosx+2sin^2x.cosx=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(2cos^2x+1\right)-cosx\left(3-2sin^2x\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(2cos^2x+1\right)-cosx\left(1+2cos^2x\right)=0\)

\(\Leftrightarrow\left(1-sinx-cosx\right)\left(2cos^2x+1\right)=0\)

\(\Leftrightarrow sinx+cosx=1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\left(ktm\right)\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)