Cho $\cos \alpha {\rm{ = }}\dfrac{3}{4};\sin \alpha  > 0$ . Tính $\cos 2\alpha ,\sin \alpha$

$\sin {225^0} = \dfrac{{ - \sqrt 2 }}{2}$

${\rm{cos22}}{{\rm{5}}^0} =  - \dfrac{{\sqrt 2 }}{2}$

$\tan {225^0} =  - 1$

$\cot {225^0} = 1$

$- 3{x^2} + x - 1 \ge 0.$

$- 3{x^2} + x - 1 > 0.$

$- 3{x^2} + x - 1 < 0.$

$3{x^2} + x - 1 \le 0.$

$\left( {3; + \,\infty } \right).$

$\left( { - \,\infty ;3} \right).$

$\left( { - \,3;3} \right).$

$\mathbb{R}.$

Chỉ $\left( {\rm{I}} \right)$ và $\left( {{\rm{II}}} \right)$.

Chỉ $\left( {\rm{I}} \right)$, $\left( {{\rm{II}}} \right)$ và $\left( {{\rm{III}}} \right)$.

Chỉ $\left( {{\rm{II}}} \right)$, $\left( {{\rm{III}}} \right)$ và $\left( {{\rm{IV}}} \right)$.

Chỉ $\left( {\rm{I}} \right)$, $\left( {{\rm{II}}} \right)$ và $\left( {{\rm{IV}}} \right)$.

$\left( {{d_1}} \right):3x + 2y = 0$

$\left( {{d_2}} \right):3x - 2y = 0$

$\left( {{d_3}} \right): - 3x + 2y - 7 = 0.$

$\left( {{d_4}} \right):6x - 4y - 14 = 0.$

$1{\rm{ rad }} = {1^0}.$

$1{\rm{ rad }} = {60^0}.$

$1{\rm{ rad }} = {180^0}.$

$1{\rm{ rad }} = {\left( {\dfrac{{180}}{\pi }} \right)^0}.$

$\pi {\rm{ rad }} = {1^0}.$

$\pi {\rm{ rad }} = {60^0}.$

$\pi {\rm{ rad }} = {180^0}.$

$\pi {\rm{ rad }} = {\left( {\dfrac{{180}}{\pi }} \right)^0}.$