1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

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

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

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

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

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

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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

3) 2sin^2 x - 3sinx + 1 = 0

Đặt t = sin x

(*) <=> 2t^2 - 3t + 1 = 0

<=> t = 1 (nhận) or t = 1/2 (nhận)

.Vs t = 1 => sinx = 1

<=> x = π/2 + k2π (k thuộc Z) (nhận)

.Vs t = 1/2 => sinx = 1/2

<=> sinx = sin π/6

<=> x = π/6 + k2π (k thuộc Z) (nhận)

Vậy ...

2) cos^2 x + cosx = 0

Đặt t = cosx

(*) <=> t^2 + t =0 <=> t = 0 (n) or t = -1 (n)

. Vs t = 0 => cosx = 0 <=> x = π/2 + kπ (loại)

.Vs t = -1 => cosx = -1 <=> x = π + k2π (nhận)

Vậy ...

1) (sin3x)/cosx + 1 = 0

ĐK: cosx + 1 ≠ 0 <=> cosx ≠ -1 <=> x ≠ π + k2π

<=> sin3x = 0

<=> 3x = kπ

<=> x = 1/3 kπ (k thuộc Z) (n)

Vậy ...

a) Dat sin x =  y

y² - 3y - 4= 0

y = -1 hoac y = 4 (loai)

voi y = -1 thi sin x = -1 => \(x=-\frac{\pi}{2}+2k\pi\)

b) Chia hai ve cho 2 ta co:

\(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin2x\)

\(cos\frac{\pi}{6}sinx+sin\frac{\pi}{6}cosx=sin2x\)

\(sin\left(x+\frac{\pi}{6}\right)=sin2x\)

\(x+\frac{\pi}{6}=2x+2k\pi\) hoac \(x+\frac{\pi}{6}=\pi-2x+2k\pi\) 

1.

\(y=\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)\Rightarrow\) tập giá trị là \(\left[-\sqrt{2};\sqrt{2}\right]\)

2. ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)

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

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow2y+y.cosx=sinx+2cosx+3\)

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

\(\Rightarrow1^2+\left(2-y\right)^2\ge\left(2y-3\right)^2\)

\(\Leftrightarrow3y^2-8y+4\le0\)

\(\Rightarrow\frac{2}{3}\le y\le2\)

4.

\(y=2\left(\frac{\sqrt{3}}{2}cos\frac{x}{2}-\frac{1}{2}sin\frac{x}{2}\right)=2cos\left(\frac{x}{2}+\frac{\pi}{6}\right)\)

\(\Rightarrow-2\le y\le2\)

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=\frac{1}{2}sin7x-\frac{\sqrt{3}}{2}cos7x\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\7x-\frac{\pi}{3}=\frac{5\pi}{6}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

6.

\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

7.

Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn

8.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)

\(sinx+cos\left(2x+\dfrac{\Omega}{3}\right)=0\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=-sinx=sin\left(-x\right)\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=cos\left(\dfrac{\Omega}{2}+x\right)\)

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

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{6}+k2\Omega\\3x=-\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)

TH1: \(x=\dfrac{5}{6}\Omega+k2\Omega\)

\(0< =x< =2\Omega\)

=>\(0< =\dfrac{5}{6}\Omega+k2\Omega< =2\Omega\)

=>\(-\dfrac{5}{6}\Omega< =k2\Omega< =\dfrac{7}{6}\Omega\)

=>\(-\dfrac{5}{6}< =2k< =\dfrac{7}{6}\)

=>-5/12<=k<=7/12

mà k nguyên

nên k=0

TH2: \(x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\)

\(0< =x< =2\Omega\)

=>\(0< =-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}< =2\Omega\)

=>\(\dfrac{5}{18}\Omega< =\dfrac{k2\Omega}{3}< =\dfrac{41}{18}\Omega\)

=>\(\dfrac{5}{18}< =\dfrac{2k}{3}< =\dfrac{41}{18}\)

=>\(\dfrac{5}{6}< =2k< =\dfrac{41}{6}\)

=>\(\dfrac{5}{12}< =k< =\dfrac{41}{12}\)

mà k nguyên

nên \(k\in\left\{1;2;3\right\}\)

=>Có 4 nghiệm thỏa mãn