1: 90<x<180

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{4}{5}\right)^2}=-\dfrac{3}{5}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{4}{5}\cdot\dfrac{-3}{5}=\dfrac{-24}{25}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{9}{25}-1=-\dfrac{7}{25}\)

\(tan2x=\dfrac{-24}{25}:\dfrac{-7}{25}=\dfrac{24}{7}\)

2: 0<x<90

=>cosx>0

=>\(cosx=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{3}{4}-1=\dfrac{6}{4}-1=\dfrac{2}{4}=\dfrac{1}{2}\)

90 độ<x<180 độ

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{12}{13}\right)^2}=-\dfrac{5}{13}\)

\(tanx=\dfrac{12}{13}:\dfrac{-5}{13}=-\dfrac{12}{5}\)

\(E=\dfrac{6\cdot\dfrac{-12}{5}+\dfrac{12}{13}}{2\cdot\dfrac{-5}{13}+\dfrac{5}{12}}=\dfrac{-\dfrac{72}{5}+\dfrac{12}{13}}{-\dfrac{10}{13}+\dfrac{5}{12}}=\dfrac{10512}{275}\)

\(\Leftrightarrow3sin^2x-2sinx.cosx-5cos^2x=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x-2tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-45^0+k180^0\\x=arctan\left(\frac{5}{3}\right)+k180^0\end{matrix}\right.\)

X= 200 - 300 = -100

X= 900 + 400 = 1300

X= 180 : 90 = 2

X= 240 : 60 = 4 

Tìm x :

x + 300 = 200

x = 200 - 300

x = -100

x - 400 = 900

x = 900 + 400

x = 1300

X x 90 = 180

 X = 180 : 90

X = 2

240 : x = 60

x = 240 : 60

x = 4

\(x+300=200\)

\(\Rightarrow x=200-300\Rightarrow x=-100\)

​\(x-400=900\Rightarrow x=900-400=500\)​

​​\(xX90=180\Rightarrow x=180:90=9\)\(;\)\(240:x=60\Rightarrow x=240:60=4\)

1) \(x+y=10\) mà \(x=y\) nên: \(x=y=\dfrac{10}{2}=5\)

2) \(2x+3y=180\) mà \(x=y\)

Ta có: \(2y+3y=180\Rightarrow5y=180\Rightarrow y=180:5=36\)

Vậy \(x=y=36\)

3) \(x+y=180\) mà \(x=y\) nên: \(x=y=\dfrac{180}{2}=90\)

4) \(3x+5y=13\) mà \(y=2x\) ta có:

\(3x+5\cdot2x=13\Rightarrow13x=13\Rightarrow x=1\)

\(y=2x=2\cdot1=2\)

Các câu còn lại bạn làm tương tự

1.

\(\Leftrightarrow2sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)

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

\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))

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

2.

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)

Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)

3. Làm tương tự câu 2

4.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(24\times\left(15+30+85-120\right)\div16\)

\(=24\times10\div16\)

\(=15\)

\(140-180\times\left(47-90+43+7\right)\)

\(=140-\left(180\times7\right)\)

\(=140-1260\)

\(=-1120\)

khó quá đi ! 

24.(15+30+85-120):16

=24.10:16

=240:16

=15

140-180.(47-90+43+7)

=140-180.7

=140-1260

=-1120