- Phương trình lượng giác giải bằng phương pháp đặ...

A \(x =  \pm \frac{{2\pi }}{3} + k\pi \)

B \(x =  \pm \frac{\pi }{3} + k\pi \)

C \(x =  \pm \frac{\pi }{6} + k\pi \)

D \(x =  \pm \frac{\pi }{6} + k2\pi \)

A \(4{t^2} - 8t + 3 = 0\)

B \(4{t^2} - 8t - 3 = 0\)

C \(4{t^2} + 8t - 5 = 0\)

D \(4{t^2} - 8t + 5 = 0\)

A \(2{t^2} + 3\sqrt 2 \,t + 2 = 0.\)

B \(4{t^2} + 3\sqrt 2 \,t + 4 = 0.\)

C \(2{t^2} + 3\sqrt 2 \,t - 2 = 0.\)

D \(4{t^2} + 3\sqrt 2 \,t - 4 = 0.\)

A \(\sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}.\)

B \(\cos \left( {x - \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}.\)

C \(\tan x = 1.\)

D \(1 + {\tan ^2}x = 0.\)

A \(\sin t = \sin 3t\)          

B \(\cos t = \sin 3t\)

C \(\sin 3t = \sin \left( { - t} \right)\)

D \(\cos t = \cos 3t\)

A \(t = 1\) hoặc \(t = \sqrt 2 \).

B \(t = 1\) hoặc \(t = \sqrt 3 \)

C \(t = 1\)

D \(t = \sqrt 3 \)

A \(P = \frac{{\sqrt 2 }}{2}\)

B \(P = 1\)

C \(P = \frac{1}{2}\)

D \(P =  - \frac{{\sqrt 2 }}{2}\)

A  \(\sin \left( {x - \frac{\pi }{4}} \right) = 0\) hoặc \(\sin \left( {x - \frac{\pi }{4}} \right) = 1\).

B \(\sin \left( {x - \frac{\pi }{4}} \right) = 0\) hoặc \(\sin \left( {x - \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\).   

C \(\sin \left( {x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\)

D \(\sin \left( {x - \frac{\pi }{4}} \right) = 0\) hoặc \(\sin \left( {x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\).

A \(\cos \left( {x + \frac{\pi }{4}} \right) =  - 1.\)

B \(\cos \left( {x + \frac{\pi }{4}} \right) = 1.\)

C \(\cos \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}.\)

D \(\cos \left( {x + \frac{\pi }{4}} \right) =  - \frac{1}{{\sqrt 2 }}.\)

A \(\sin 2x =  - \frac{1}{2}.\)

B \(\sin 2x =  - \frac{{\sqrt 2 }}{2}.\)

C \(\sin 2x = \frac{1}{2}.\)

D \(\sin 2x = \frac{{\sqrt 2 }}{2}.\)

A \(2\pi \)

B \(4\pi \)

C \(3\pi \)

D \(\pi \)

A 1

B 2

C 3

D 4

A \(\pi  + k\pi \)

B \(\pi  + k2\pi \)

C \(\frac{\pi }{2} + k2\pi \)

D \(\frac{\pi }{4} + k\pi \)

A \(x = \pi  + \frac{{k\pi }}{4}(k \in Z)\)

B \(\frac{\pi }{2} + \frac{{k\pi }}{2}(k \in Z)\)

C \(\frac{\pi }{2} + k\pi (k \in Z)\)

D \(\frac{\pi }{4} + \frac{{k\pi }}{2}\left( {k \in Z} \right)\)

A \(x =  \pm \alpha  + k\pi \) với \(cos\alpha  = \frac{{1 - \sqrt 5 }}{2}\;\;\left( {k \in Z} \right)\)        

B \(x =  \pm \alpha  + k2\pi \) với \(cos\alpha  = \frac{{1 - \sqrt 5 }}{2}\;\;\left( {k \in Z} \right)\)     

C \(x =  \pm \alpha  + \frac{{k\pi }}{2}\) với \(cos\alpha  = \frac{{1 - \sqrt 5 }}{2}\;\;\left( {k \in Z} \right)\)

D \(x =  \pm \alpha  + \frac{{k\pi }}{4}\) với \(cos\alpha  = \frac{{1 - \sqrt 5 }}{2}\;\;\left( {k \in Z} \right)\)

A \(\left[ \begin{array}{l}x = k2\pi \left( {k \in Z} \right)\\x = \frac{\pi }{4} + m2\pi \left( {m \in Z} \right)\end{array} \right.\)

B \(\left[ \begin{array}{l}x = k2\pi \;\left( {k \in Z} \right)\\x = \frac{{3\pi }}{2} + m2\pi \;\left( {m \in Z} \right)\end{array} \right.\)

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

D \(\left[ \begin{array}{l}x = k2\pi \left( {k \in Z} \right)\\x = \frac{\pi }{2} + m2\pi \;\left( {m \in Z} \right)\end{array} \right.\)

A \(\left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = m\pi \end{array} \right.\)

B \(\left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = m2\pi \end{array} \right.\)

C \(\left[ \begin{array}{l}x = \frac{{3\pi }}{4} + k\pi \\x = m\frac{\pi }{2}\end{array} \right.\)

D \(\left[ \begin{array}{l}x = \frac{{3\pi }}{2} + k2\pi \\x = \left( {2m + 1} \right)\pi \end{array} \right.\)

A \({x_0} \in \left( {0;\frac{\pi }{{12}}} \right).\)

B \({x_0} \in \left[ {\frac{\pi }{{12}};\frac{\pi }{6}} \right].\)

C \({x_0} \in \left( {\frac{\pi }{6};\frac{\pi }{3}} \right].\)

D \({x_0} \in \left( {\frac{\pi }{3};\frac{\pi }{2}} \right].\)