a: \(A=2\left(\sin^6x+cos^6x\right)-3\cdot\left(\sin^4x+cos^4x\right)\)

\(=2\cdot\left\lbrack\left(\sin^2x+cos^2x\right)^3-3\cdot\sin^2x\cdot cos^2x\cdot\left(\sin^2x+cos^2x\right)\right\rbrack-3\cdot\left\lbrack\left(sin^2x+cos^2x\right)^2-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2-6\cdot\sin^2x\cdot cos^2x-3+6\cdot\sin^2x\cdot cos^2x\)

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

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

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

\(=\sin^2x+cos^2x=1\)

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

\(=cos^2x\left(\frac{1}{\sin^2x}-1\right)\cdot\frac{\sin^2x}{cos^2x}+\frac{\sin x\cdot cosx\cdot\sin x}{cosx}\)

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

\(\Leftrightarrow\dfrac{2}{\sin2x}=2\)

\(\Leftrightarrow\sin2x=1\)

\(\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\)

hay \(x=\dfrac{\Pi}{4}+k\Pi\)

Lời giải:
$\tan x +\cot x=2$. Mà $\tan x\cot x =1$

$\Rightarrow \tan x = \cot x =1$

$\Rightarrow x=45^0$

$\Rightarrow A=\sin x\cos x =\sin 45^0.\cos 45^0=\frac{1}{2}$

$B=\sin x+\cos x= \sin 45^0+\cos 45^0=\sqrt{2}$

a: \(\Leftrightarrow1-cos^4x-cos^2x=1\)

\(\Leftrightarrow cos^2x\left(cos^2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=k2\Pi\\x=\Pi+k2\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow3\left(1+\tan^2x\right)+2\sqrt{3}tanx-6=0\)

\(\Leftrightarrow3\cdot tan^2x+2\sqrt{3}\cdot tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\dfrac{\sqrt{3}}{3}\\tanx=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{6}+k\Pi\\x=-\dfrac{\Pi}{3}+k\Pi\end{matrix}\right.\)

Ta có: *nếu x =  45 °  thì tgx = cotgx, suy ra: tgx – cotgx = 0

*nếu x <  45 °  thì cotgx = tg( 90 °  – x)

Vì x <  45 °  nên  90 °  – x >  45 ° , suy ra: tgx < tg( 90 °  – x)

Vậy tgx – cotgx < 0

*nếu x > 45 °  thì cotgx = tg( 90 ° – x)

Vì x >  45 °  nên  90 °  – x <  45 ° , suy ra: tgx > tg( 90 °  – x)

Vậy tgx – cotgx > 0.

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: