a) <=> 4sinxcosx -(2cos²x-1)=7sinx+2cosx-4

<=> 2cos²x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos²x-2cosx+2=0 

pt trên vn

b) <=> 2sinxcosx-1+2sin²x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin²x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a.

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

b.

\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

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

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

c.

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

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

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

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

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

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

\(\Leftrightarrow...\)

\(\left(2cosx+1\right)\left(3cos2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\3cos2x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cos2x=\dfrac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)

Tương đương cosx = cos120⁰

a,Pt \(\Leftrightarrow cosx-sinx=\dfrac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=-\dfrac{\pi}{4}-arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) ,\(k\in Z\)

b) Pt \(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)

Đặt \(cosa=\dfrac{4}{5}\Rightarrow sina=\dfrac{3}{5}\)

Pttt:\(cosx.cosa-sina.sinx=\dfrac{3}{5}\)

\(\Leftrightarrow cos\left(x+a\right)=\dfrac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-a+arc.cos\left(\dfrac{3}{5}\right)+2k\pi\\x=-a-arc.cos\left(\dfrac{3}{5}\right)+2k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}.sin3x=1\)

Đặt \(cosa=\dfrac{3}{5}\Rightarrow sina=\dfrac{4}{5}\)

Pttt:\(cos3x.cosa+sin3a.sina=1\)

\(\Leftrightarrow cos\left(3x-a\right)=1\)

\(\Leftrightarrow x=\dfrac{a}{3}+\dfrac{k2\pi}{3}\)(\(k\in Z\))

Vậy...

1)\(1+2sinx=2cosx\)

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

\(\Leftrightarrow\left(cosx-sinx\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow cosx^2+sinx^2-2cosxsinx=\dfrac{1}{4}\)

\(\Leftrightarrow1-2cosxsinx=\dfrac{1}{4}\)

\(\Leftrightarrow2cosxsinx=\dfrac{3}{4}\)

\(\Leftrightarrow sin2x=\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x=arcsin\dfrac{3}{8}+k\pi\\x=\pi-arcsin\dfrac{3}{8}+k\pi\end{matrix}\right.\) \(\left(K\in Z\right)\)

b) \(4cosx-3sinx=3\)

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

Đặt \(cosa=\dfrac{3}{5},sina=\dfrac{4}{5}\)

Khi đó:

\(sinacosx-cosasinx=\dfrac{3}{5}\)

\(\Leftrightarrow sin\left(a-x\right)=\dfrac{3}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-x=arcsin\dfrac{3}{5}+k2\pi\\a-x=\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=a-arcsin\dfrac{3}{5}+k2\pi\\x=a-\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

3)\(3cos3x+4sin3x=5\)

\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}sin3x=1\)

Đặt \(sina=\dfrac{3}{5},cosa=\dfrac{4}{5}\)

khi đó: \(sinacos3x+cosasin3x=1\)

\(\Leftrightarrow sin\left(a+3x\right)=\dfrac{\pi}{2}\)

\(\Leftrightarrow3x=\dfrac{\pi}{2}-a+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}-\dfrac{1}{3}a+k\dfrac{2}{3}\pi\),\(k\in Z\)

Chúc bạn học tốt^^

3 cos 2 x   -   2 sin 2 x   +   sin 2 x   =   1

Với cosx = 0 ta thấy hai vế đều bằng 1. Vậy phương trình có nghiệm x = 0,5π + kπ, k ∈ Z

Trường hợp cosx ≠ 0, chia hai vế cho cos2x ta được:

3   -   4 tan x   +   tan 2 x   =   1   +   tan 2 x     ⇔   4 tan x   =   2     ⇔   tan x   =   0 , 5     ⇔   x   =   a r c tan   0 , 5   +   k π ,   k   ∈   Z

Vậy nghiệm của phương trình là

x = 0,5π + kπ, k ∈ Z

và x = arctan 0,5 + kπ, k ∈ Z

cos (2x) + cos (4x) + cos (6x) + cos (8x)=cos (8x) + cos (6x) + cos (4x) + cos (2x)

cos (8x) + cos (6x) + cos (4x) +cos (2x)=0