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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

Vậy ... 

ĐK: 1 - sin2x ≠ 0 <=> sin2x ≠ 1 <=> x ≠ π/4 + kπ

(*) <=> 2cos2x = 0

<=> cos2x = 0

<=> 2x = π/2 + kπ

<=> x = π/4 + kπ/2 (k thuộc Z)

So vs ĐK ta đc: x = 3π/4 + kπ (k thuộc Z)

Vậy ...

1   +   sin x   -   cos x   -   sin 2 x   +   2 cos 2 x   =   0   ( 1 )     T a   c ó :     1   -   sin 2 x   =   sin x   -   cos x 2     ⇔   2 cos 2 x   =   2 ( cos 2 x   -   sin 2 x )   =   - 2 ( sin x   -   cos x ) ( sin x   +   cos x )     V ậ y   ( 1 )   ⇔   ( sin x   -   cos x ) ( 1   +   sin x   -   cos x   -   2 sin x   -   2 cos x )   =   0     ⇔   ( sin x   -   cos x ) ( 1   -   sin x   -   3 cos x )   =   0

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

d.

\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=0\)

\(\Leftrightarrow sin^2x-cos^2x=0\)

\(\Leftrightarrow-cos2x=0\)

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

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

e. Đề thiếu

f.

\(\Leftrightarrow sin2x=\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)\)

\(\Leftrightarrow sin2x=cos^2\frac{x}{2}-sin^2\frac{x}{2}\)

\(\Leftrightarrow sin2x=cosx\)

\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)

b.

\(\Leftrightarrow sin2x=1\)

\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)

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

c.

\(\Leftrightarrow2sin2x.cos2x=-1\)

\(\Leftrightarrow sin4x=-1\)

\(\Leftrightarrow4x=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

Thấy cosx= 0 là nghiệm của phương trình => \(x=\dfrac{\pi}{2}+k\pi\)

Xét cosx khác 0, chia cả 2 vế cho cos^2 x

\(\Leftrightarrow\tan^2x-\sqrt{3}\tan x+2=1+\tan^2x\)

\(\Leftrightarrow\tan x=\dfrac{\sqrt{3}}{3}\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)