## Bài 1: A/sin^2(2x)=1/2 B/cot^2(x/2)=1/3 C/tan(π/12+12x)=-căn3

Question

Bài 1:
A/sin^2(2x)=1/2
B/cot^2(x/2)=1/3
C/tan(π/12+12x)=-căn3

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9 months 2020-10-31T22:15:42+00:00 1 Answers 57 views 0

a. $$\left[ \begin{array}{l} x = \dfrac{\pi }{{12}} + k\pi \\ x = \dfrac{{5\pi }}{{12}} + k\pi \\ x = – \dfrac{\pi }{{12}} + k\pi \\ x = \dfrac{{7\pi }}{{12}} + k\pi \end{array} \right.$$
$$\begin{array}{l} a.{\sin ^2}\left( {2x} \right) = \dfrac{1}{2}\\ \to \left[ \begin{array}{l} \sin 2x = \dfrac{1}{2}\\ \sin 2x = – \dfrac{1}{2} \end{array} \right.\\ \to \left[ \begin{array}{l} 2x = \dfrac{\pi }{6} + k2\pi \\ 2x = \dfrac{{5\pi }}{6} + k2\pi \\ 2x = – \dfrac{\pi }{6} + k2\pi \\ 2x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l} x = \dfrac{\pi }{{12}} + k\pi \\ x = \dfrac{{5\pi }}{{12}} + k\pi \\ x = – \dfrac{\pi }{{12}} + k\pi \\ x = \dfrac{{7\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in Z} \right)\\ b.DK:\sin \left( {\dfrac{x}{2}} \right) \ne 0\\ {\cot ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{3}\\ \to {\tan ^2}\dfrac{x}{2} = 3\\ \to \left[ \begin{array}{l} \tan \dfrac{x}{2} = \sqrt 3 \\ \tan \dfrac{x}{2} = – \sqrt 3 \end{array} \right.\\ \to \left[ \begin{array}{l} \dfrac{x}{2} = \dfrac{\pi }{3} + k\pi \\ \dfrac{x}{2} = – \dfrac{\pi }{3} + k\pi \end{array} \right.\\ \to \left[ \begin{array}{l} x = \dfrac{{2\pi }}{3} + k2\pi \\ x = – \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\\ c.DK:\cos \left( {\dfrac{\pi }{{12}} + 12x} \right) \ne 0\\ \tan \left( {\dfrac{\pi }{{12}} + 12x} \right) = – \sqrt 3 \\ \to \dfrac{\pi }{{12}} + 12x = – \dfrac{\pi }{3} + k\pi \\ \to 12x = – \dfrac{{5\pi }}{{12}} + k\pi \\ \to x = – 5\pi + k12\pi \left( {k \in Z} \right) \end{array}$$