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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a/

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

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

b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)

\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)

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

\(\Leftrightarrow cosx=1-2cos^2x\)

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

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

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

c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)

\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)

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

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

ĐKXĐ: sin2x≠\(\dfrac{\sqrt{3}}{2}\)

(1) ⇔ sin2x + cosx - \(\sqrt{3}\) ( cos2x + sinx) = 2sin2x - \(\sqrt{3}\)

⇔cosx - \(\sqrt{3}\) sinx = \(\sqrt{3}\) cos2x + sin2x +\(\sqrt{3}\)

⇔\(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}\)

⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x+\dfrac{\Pi}{3}\right)-sin\dfrac{\Pi}{3}\)

⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2cos\left(x+\dfrac{\Pi}{3}\right)sinx\)

⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2sin\left(\dfrac{\Pi}{6}-x\right)sinx\)

⇔\(sin\left(\dfrac{\Pi}{6}-x\right)\left(2sinx-1\right)=0\)

Đến đây tự giải tiếp nha nhớ đối chiếu đk.

b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)

ĐKXĐ: sinx≠0 và cosx≠1

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

⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin²x

⇔2cos³x - cos²x - 2cosx +1 = 0

⇔ (cosx-1)(cosx+1)(2cosx-1)=0

hộ vs ae ơi

ĐKXĐ: \(sinx\ne0\)

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

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

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

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

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

\(\Leftrightarrow sinx+cosx=1\)

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

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

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

câu 1

⇒ \(\dfrac{cosx}{sinx}\) - \(\dfrac{sinx}{cosx}\) -\(\dfrac{2cos4x}{2sinxcosx}\) =0

⇔ \(\dfrac{cos^2x-sin^2x}{sinx.cosx}\) -\(\dfrac{cos4x}{sinx.cosx}\)= 0

⇔ \(\dfrac{cos2x-cos4x}{sinx.cosx}\) = 0

\(\left[{}\begin{matrix}cos2x=cos4x\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4x+k2\pi\\2x=-4x+k2\pi\\2x=k\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-k\pi\\x=\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\) (k∈ Z)

câu 2 dùng công thức biến đổi tích thành tổng thành cos 4x + cos 2x sau đó phương trình trở thành sin x - cos 4x=0

ĐKXĐ: ...

a/ \(\frac{sin2x}{cos2x}+\frac{cosx}{sinx}=8cos^2x\)

\(\Leftrightarrow sin2x.sinx+cos2x.cosx=8cos^2x.sinx.cos2x\)

\(\Leftrightarrow cosx=4sin2x.cos2x.cosx\)

\(\Leftrightarrow cosx=2sin4x.cosx\)

\(\Leftrightarrow cosx\left(2sin4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin4x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

b/ \(\frac{cosx}{sinx}-\frac{sinx}{cosx}+4sin2x=\frac{1}{sinx.cosx}\)

\(\Leftrightarrow cos^2x-sin^2x+4sin2x.sinx.cosx=1\)

\(\Leftrightarrow cos2x+2sin^22x=1\)

\(\Leftrightarrow cos2x+2\left(1-cos^22x\right)=1\)

\(\Leftrightarrow-2cos^22x+cos2x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

1c/

\(5sinx-2=3\left(1-sinx\right)\frac{sin^2x}{1-sin^2x}\)

\(\Leftrightarrow5sinx-2=\frac{3sin^2x}{1+sinx}\)

\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)

\(\Leftrightarrow5sin^2x+3sinx-2=3sin^2x\)

\(\Leftrightarrow2sin^2x+3sinx-2=0\)

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

Bài 2:

a/ \(\Leftrightarrow\frac{\left(m+1\right)\left(1-cos2x\right)}{2}-sin2x+cos2x=0\)

\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)

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

\(4+\left(m-1\right)^2\ge\left(m+1\right)^2\)

\(\Leftrightarrow4m\le4\Rightarrow m\le1\)

Bài 2:

b/ \(\Leftrightarrow1-cos2x+msin2x=2m\)

\(\Leftrightarrow msin2x-cos2x=2m-1\)

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

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

\(\Leftrightarrow3m^2-4m\le0\)

\(\Rightarrow0\le m\le\frac{4}{3}\)

c/ Với \(cosx=0\) không phải là nghiệm

Với \(cosx\ne0\), chia 2 vế cho \(cos^2x\) ta được:

\(tan^2x-4tanx+m-2=0\)

Đặt \(tanx=t\Rightarrow t\in\left[0;1\right]\)

Phương trình trở thành: \(t^2-4t+m-2=0\)

\(\Leftrightarrow f\left(t\right)=t^2-4t-2=-m\)

Dựa vào đồ thị hàm \(f\left(t\right)=t^2-4t-2\), để \(y=-m\) cắt \(y=f\left(t\right)\) với \(t\in\left[0;1\right]\) \(\Rightarrow-5\le-m\le-2\)

\(\Rightarrow2\le m\le5\)

\(\text{c) }sin3x-\sqrt{3}cos3x=2cos5x\\ \Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=cos5x\\ \Leftrightarrow sin\frac{\pi}{6}\cdot sin3x-cos\frac{\pi}{6}\cdot cos3x=cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=-cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=cos\left(\pi-5x\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=\pi-5x+m2\pi\\3x+\frac{\pi}{6}=5x-\pi+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{48}+\frac{m\pi}{4}\\x=\frac{7\pi}{12}-n\pi\end{matrix}\right.\)

\(d\text{) }sinx\left(sinx+2cosx\right)=2\\ \Leftrightarrow cos^2x+\left(sinx-cosx\right)^2=0\\ \Leftrightarrow cosx=sinx=0\left(VN\right)\)

\(e\text{) }\sqrt{3}\left(sin2x+cos7x\right)=sin7x-cos2x\\ \Leftrightarrow\sqrt{3}sin2x+cos2x=sin7x-\sqrt{3}cos7x\\ \Leftrightarrow sin2x\cdot\frac{\sqrt{3}}{2}+cos2x\cdot\frac{1}{2}=sin7x\cdot\frac{1}{2}-cos7x\cdot\frac{\sqrt{3}}{2}\\ \Leftrightarrow sin2x\cdot cos\frac{\pi}{3}+cos2x\cdot sin\frac{\pi}{3}=sin7x\cdot cos\frac{\pi}{3}-cos7x\cdot sin\frac{\pi}{3}\\ \Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=7x-\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=\frac{4\pi}{3}-7x+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-m2\pi}{5}\\x=\frac{5\pi}{27}+\frac{n2\pi}{9}\end{matrix}\right.\)

\(\text{a) }\sqrt{3}sin2x-cos2x+1=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=-\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2x=\frac{1}{2}\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+m\pi\\x=n\pi\end{matrix}\right.\)

\(\text{b) }pt\Leftrightarrow sin4x=\frac{1-4cosx}{3}\\ \Leftrightarrow sin^24x+cos^24x=\left(\frac{1-cos4x}{3}\right)^2+cos^24x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{arccos\left(-\frac{4}{5}\right)}{4}+\frac{k\pi}{2}\end{matrix}\right.\)