27 câu Trắc nghiệm Đại số và Giải tích 11 Bài 2 (Có đáp án): Phương trình lượng giác cơ bản

π/2+k4π, k ∈ Z.

π/2+kπ, k ∈ Z.

π/4+k2π, k ∈ Z.

π/4+kπ, k ∈ Z.

π/2+k2π, k ∈ Z.

3π/2+k2π, k ∈ Z.

3π/2+k3π, k ∈ Z.

kπ, k ∈ Z.

$\left\{\frac{\mathrm{\pi }}{6};\frac{11\mathrm{\pi }}{6}\right\}$

$\left\{\frac{2\mathrm{\pi }}{3};\frac{4\mathrm{\pi }}{3}\right\}$

$\left\{\frac{\mathrm{\pi }}{3};\frac{5\mathrm{\pi }}{3}\right\}$

$\left\{\frac{5\mathrm{\pi }}{6};\frac{7\mathrm{\pi }}{6}\right\}$

x = kπ, k ∈ Z.

π+k2π, k ∈ Z.

π/2+kπ, k ∈ Z.

±π/6+kπ, k ∈ Z.

x = 2π + k4π, k ∈ Z

x = k2π, k ∈ Z.

x = π + k2π, k ∈ Z.

x = 2π + kπ, k ∈ Z.

x = kπ, k ∈ Z.

x =kπ/2, k ∈ Z.

x =kπ/3, k ∈ Z.

x =kπ/4, k ∈ Z.

x = π/4 + kπ, k ∈ Z.

x = 3π/4 + kπ, k ∈ Z.

x = kπ, k ∈ Z.

x = k2π, k ∈ Z.

x = - π/4 + kπ, k ∈ Z.

x = π/4 + kπ, k ∈ Z.

x = - π/4 + k2π, k ∈ Z.

x = π/4 + k2π, k ∈ Z.

$\left\{\frac{\mathrm{\pi }}{2}\right\}$

$\left\{\frac{\mathrm{\pi }}{2};\frac{3\mathrm{\pi }}{2}\right\}$

$\left\{0;\mathrm{\pi }\right\}$

$\left\{0;\frac{\mathrm{\pi }}{2};\mathrm{\pi }\right\}$

$\left\{\frac{7\mathrm{\pi }}{6};\frac{11\mathrm{\pi }}{6}\right\}$

$\left\{\frac{\mathrm{\pi }}{2};\frac{7\mathrm{\pi }}{6};\frac{11\mathrm{\pi }}{6}\right\}$

$\left\{\frac{5\mathrm{\pi }}{6};\frac{7\mathrm{\pi }}{6}\right\}$

$\left\{\frac{\mathrm{\pi }}{6};\frac{7\mathrm{\pi }}{6};\frac{5\mathrm{\pi }}{6}\right\}$

0

2

1

3

1

0

2

3

{2π/3}

{3π/2}

{π/3; 2π/3}

{π/2; 3π/2}

{0;π}

{0;π/2}

{π/2; 3π/2}

{0; 3π/2}

x = kπ, k ∈ Z.

x = π + k2π, k ∈ Z.

π/2+kπ, k ∈ Z.

π/4+kπ, k ∈ Z.

x = π/8+k π/4, k ∈ Z.

x = π/4+k π/2, k ∈ Z.

x = π/6+k π/3, k ∈ Z.

x = π/2+kπ, k ∈ Z.

{π/2+kπ, k ∈ Z}

{π/2+k2π, k ∈ Z}

∅

{-π/2+k2π, k ∈ Z}

{π/3+k2π, k ∈ Z}

{±π/3+k2π, k ∈ Z}

{±π/3+k2π, - π/2+k2π, k ∈ Z}

{- π/2+k2π, k ∈ Z}

{π/4;3π/4}

{π/4}

{3π/4}

{π/6;π/4;3π/4}

{π/4+kπ, k ∈ Z}

{π/4+kπ/2, k ∈ Z}

{π/4+k2π, k ∈ Z}

{kπ, k ∈ Z}

x = ±5π/3 +k4π

x = ±5π/6 +k2π

x = ±5π/6 +k4π

x = ±5π/3 +kπ

\[\frac{\pi }{3} + k\pi ,\,\,k \in \mathbb{Z}\]

\[\frac{{ - \pi }}{3} + k2\pi ,\,\,k \in \mathbb{Z}\]

\[\frac{\pi }{6} + k\pi ,\,\,k \in \mathbb{Z}\]

\[\frac{{ - \pi }}{3} + k\pi ,\,\,k \in \mathbb{Z}\]

$\sin x=-1 \iff x=-\frac{\mathrm{\pi }}{2}+k2\mathrm{\pi }$

$\sin x=0 \iff x=k\mathrm{\pi }$

$\sin x=0 \iff x=k2\mathrm{\pi }$

$\sin x=1 \iff x=\frac{\mathrm{\pi }}{2}+k2\mathrm{\pi }$

$x=k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\mathrm{k}2\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$

$x=k\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\mathrm{k\pi }\pm \frac{\mathrm{\pi }}{6}$

$x=k2\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\mathrm{k}2\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$

$x=\pm \frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi }$

x = π/2

x = 0

x = π

x = - π/2

1

4

5

2