\(\tan x=\frac{\sin x}{\cos x}=\frac{3}{5}\Rightarrow\sin x=\frac{3}{5}\cos x\)

\(\Rightarrow N=\frac{\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\sin x.\cos x}{\left(\sin x-\cos x\right)\left(\sin x+\cos x\right)}\)

\(=\frac{\frac{3}{5}.\cos^2x}{\left(\frac{3}{5}\cos x-\cos x\right)\left(\frac{3}{5}\cos x+\cos x\right)}=\frac{\frac{3}{5}\cos^2x}{\frac{-16}{25}.\cos^2x}=\frac{-15}{16}\)

a) Dùng bảng lượng giác sinx = 0,2368 => x ≈ 13o42'

- Cách nhấn máy tính:

b) x ≈ 51o31'

- Cách nhấn máy tính:

c) x ≈ 65o6'

- Cách nhấn máy tính:

d) x ≈ 17o6'

- Cách nhấn máy tính:

$\sin x=0,6\\\Leftrightarrow \sin^2 x=0,36\\\Rightarrow \cos^2 x=0,64\\\Leftrightarrow \cos x=0,8(x>0)$

Cảm ơn ạ 

1.

\(sin^2x+cos^2x=1\Rightarrow\left(\dfrac{1}{4}\right)^2+cos^2x=1\)

\(\Rightarrow cos^2x=\dfrac{15}{16}\Rightarrow cosx=\dfrac{\sqrt{15}}{4}\)

2.

\(tanx=\dfrac{1}{3}\Rightarrow tan^2x=\dfrac{1}{9}\Rightarrow\dfrac{sin^2x}{cos^2x}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{sin^2x}{1-sin^2x}=\dfrac{1}{9}\Rightarrow9sin^2x=1-sin^2x\)

\(\Rightarrow sin^2x=\dfrac{1}{10}\Rightarrow sinx=\dfrac{\sqrt{10}}{10}\)

\(x=45^0\)

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

Ta có: 

\(\left\{{}\begin{matrix}tanx=3\\sin^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\9cos^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cos^2x=\dfrac{1}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cosx=\pm\dfrac{1}{\sqrt{10}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=\dfrac{3}{\sqrt{10}}\\cosx=\dfrac{1}{\sqrt{10}}\end{matrix}\right.\\\left\{{}\begin{matrix}sinx=-\dfrac{3}{\sqrt{10}}\\cosx=-\dfrac{1}{\sqrt{10}}\end{matrix}\right.\end{matrix}\right.\)