15 câu Trắc nghiệm Một số phương trình lượng giác thường gặp có đáp án (Thông hiểu)

$x=\pm \frac{2\pi }{3}+k2\pi \left(k\in Z\right)$

$x=\pm \frac{\pi }{2}+k2\pi \left(k\in Z\right)$

$x=\pm \frac{\pi }{3}+k2\pi \left(k\in Z\right)$

$x=\frac{2\pi }{3}+k2\pi \left(k\in Z\right)$

$\frac{3\pi }{5}$

$\frac{29\pi }{30}$

$\frac{5\pi }{6}$

$\frac{23\pi }{30}$

$\left[\begin{array}{l}x=-\frac{\pi }{6}+\frac{k2\pi }{9}\\ x=\frac{7\pi }{6}+\frac{k2\pi }{9}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=-\frac{\pi }{9}+\frac{k2\pi }{9}\\ x=\frac{7\pi }{9}+\frac{k2\pi }{9}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=-\frac{\pi }{12}+\frac{k2\pi }{9}\\ x=\frac{7\pi }{12}+\frac{k2\pi }{9}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=-\frac{\pi }{54}+\frac{k2\pi }{9}\\ x=\frac{\pi }{18}+\frac{k2\pi }{9}\end{array}\right.(k\in Z)$

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Vô nghiệm

$x=k\pi \left(k\in \mathbb{Z}\right)$

$x=\frac{\mathrm{\pi }}{2}+k\pi \left(k\in \mathbb{Z}\right)$

$x=\frac{\pi }{6}+k2\pi \left(k\in \mathbb{Z}\right)$

Tất cả đều đúng

$\left[\begin{array}{l}x=\frac{\pi }{6}+k2\pi \\ x=\frac{\pi }{3}+k2\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{5\pi }{6}+k\pi \\ x=\frac{\pi }{3}+k2\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{5\pi }{6}+k2\pi \\ x=\frac{\pi }{3}+k2\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{5\pi }{6}+k\pi \\ x=\frac{\pi }{3}+k\pi \end{array}\right.(k\in Z)$

Có 1 họ nghiệm

Có 2 họ nghiệm

Vô nghiệm

Có 1 nghiệm duy nhất

$\left[\begin{array}{l}x=-\frac{\pi }{18}+\frac{k2\pi }{3}\\ x=\frac{3\pi }{10}+\frac{k2\pi }{5}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{18}+\frac{k2\pi }{3}\\ x=-\frac{3\pi }{10}+\frac{k2\pi }{5}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=-\frac{\pi }{18}+\frac{k2\pi }{5}\\ x=\frac{3\pi }{10}+\frac{k2\pi }{3}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{18}+\frac{k2\pi }{3}\\ x=\frac{3\pi }{10}+\frac{k2\pi }{5}\end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{2}+k\pi \\ x=\frac{\pi }{6}+k\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{4}+k\pi \\ x=\frac{\pi }{3}+k\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{8}+k2\pi \\ x=\frac{\pi }{12}+k2\pi \end{array}\right.(k\in Z)$

$\left[\begin{array}{l}x=\frac{\pi }{8}+k\pi \\ x=\frac{\pi }{12}+k\pi \end{array}\right.(k\in Z)$

$x=\frac{\pi }{2}+k\pi \left(k\in Z\right)$

$\left[\begin{array}{l}x=\frac{\pi }{4}+k\pi \\ x=arctan(-2)+k\pi \end{array}\right.(k\in Z)$

$x=-\frac{\pi }{4}+k\pi \left(k\in Z\right)$

Vô nghiệm

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$\left[\begin{array}{c}x=\frac{\mathrm{\pi }}{3}+k\mathrm{\pi }\\ x=\frac{5\mathrm{\pi }}{3}+k\mathrm{\pi }\end{array}\right.$

$\left[\begin{array}{c}x=\frac{\mathrm{\pi }}{4}+k\mathrm{\pi }\\ x=5\mathrm{\pi }+k\mathrm{\pi }\end{array}\right.$

$\left[\begin{array}{c}x=\frac{\mathrm{\pi }}{6}+k\mathrm{\pi }\\ x=\frac{5\mathrm{\pi }}{4}+k\mathrm{\pi }\end{array}\right.$

$\left[\begin{array}{c}x=\frac{\mathrm{\pi }}{12}+k\mathrm{\pi }\\ x=\frac{5\mathrm{\pi }}{12}+k\mathrm{\pi }\end{array}\right.$

$\mathrm{x}=\frac{\mathrm{\pi }}{3}+\mathrm{k\pi }, \mathrm{x}=\mathrm{k\pi }, \mathrm{x}=\frac{\mathrm{k\pi }}{5}$

$\mathrm{x}=\frac{\mathrm{\pi }}{2}+\mathrm{k\pi }, \mathrm{x}=k2\pi , \mathrm{x}=\frac{2\mathrm{k\pi }}{5}$

$\mathrm{x}=\frac{\mathrm{\pi }}{2}+2\mathrm{k\pi }, \mathrm{x}=\frac{k2\pi }{3}$

$\mathrm{x}=\pm \frac{\mathrm{\pi }}{2}+\mathrm{k\pi }, \mathrm{x}=k2\pi$

$x=\frac{\mathrm{\pi }}{9}+\frac{k2\mathrm{\pi }}{3} \left(k\in Z\right)$

$x=\frac{\mathrm{\pi }}{18}+\frac{k\mathrm{\pi }}{6} \left(k\in Z\right)$

$x=\pm \frac{\mathrm{\pi }}{6}+\frac{k\mathrm{\pi }}{2} \left(k\in Z\right)$

$x=\frac{\mathrm{\pi }}{18}+\frac{k2\mathrm{\pi }}{3}; x=-\frac{\mathrm{\pi }}{6}+\frac{k\mathrm{\pi }}{2} \left(k\in Z\right)$