\(A=\frac{1-2sina.cosa}{sin^2a-cos^2a}=\frac{sin^2a+cos^2a-2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina-cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina-cosa}{sina+cosa}\)

b/ \(A=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}=-\frac{1}{2}\)

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

cos²x + sin²x=1

=>sin²x=1-cos²x=0.75

=>sinx=\(\pm\)\(\sqrt{3}\)/2

A= \(\dfrac{0,5+2.0,75}{0,5^2\pm\dfrac{\sqrt{3}}{2}}\)= \(\dfrac{-8\pm16\sqrt{3}}{11}\)

1/

\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)

\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)

2/

\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)

\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)

3/

\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)

\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)

\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)

\(=-1+2cos2a+1-cos2a=cos2a\)

Sửa đề: \(2\cdot sin\left(180-a\right)\cdot cota-cos\left(180-a\right)\cdot tana+cot\left(180-a\right)\)

\(=2\cdot sina\cdot cota+cosa\cdot tana+\dfrac{cos\left(180-a\right)}{sin\left(180-a\right)}\)

\(=2\cdot sina\cdot\dfrac{cosa}{sina}+cosa\cdot\dfrac{sina}{cosa}+\dfrac{-cosa}{sina}\)

\(=2cosa+sina-tana\)

a)

\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)

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

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)

c)

\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)

\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)

d)

\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)

\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)

f)

\(\cos 2x-\cos 4x=0\)

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

\(\Rightarrow \left[\begin{matrix} x=k\pi\\ x=\frac{k}{3}\pi \end{matrix}\right.\) ( k nguyên)

b,e,g bạn xem lại đề, đơn vị không thống nhất.

\(\dfrac{3sin\alpha-4cos\alpha}{2sin\alpha+3cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{3tan\alpha-4}{2tan\alpha+3}\)

Biết tanα=\(-\dfrac{1}{4}\) nên ta có:

\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)

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 \(x+\frac{\pi}{6}=\pi-2x+2k\pi\)