Giải câu 4 bài: Ôn tập chương I

a) \(\sin (x + 1) = {2 \over 3}\) (1)

(1) \(\eqalign{
& \sin (x + 1) = {2 \over 3} \cr 
& \Leftrightarrow \left[ \matrix{
x + 1 = \arcsin {2 \over 3} + k2\pi \hfill \cr 
x + 1 = \pi - \arcsin {2 \over 3} + k2\pi \hfill \cr} \right. \cr 
& \Leftrightarrow \left[ \matrix{
x = - 1 + \arcsin {2 \over 3} + k2\pi \hfill \cr 
x = - 1 + \pi - \arcsin {2 \over 3} + k2\pi \hfill \cr} \right.;k \in Z \cr} \)

b) \({\sin ^2}2x = {1 \over 2}\) (2)

(2) \(\eqalign{
& {\sin ^2}2x = {1 \over 2} \Leftrightarrow {{1 - \cos 4x} \over 2} = {1 \over 2} \cr 
& \Leftrightarrow \cos 4x = 0 \Leftrightarrow 4x = {\pi \over 2} + k\pi \cr 
& \Leftrightarrow x = {\pi \over 8} + k{\pi \over 4},k \in Z \cr} \)

c) \({\cot ^2}{x \over 2} = {1 \over 3}\) 

\(\eqalign{
& {\cot ^2}{x \over 2} = {1 \over 3} \Leftrightarrow \left[ \matrix{
\cot {x \over 2} = {{\sqrt 3 } \over 3}(1) \hfill \cr 
\cot {x \over 2} = - {{\sqrt 3 } \over 3}(2) \hfill \cr} \right. \cr 
& (1) \Leftrightarrow \cot {x \over 2} = \cot {\pi \over 3} \Leftrightarrow {x \over 2} = {\pi \over 3} + k\pi \cr 
& \Leftrightarrow x = {{2\pi } \over 3} + k2\pi ,k \in z \cr 
& (2) \Leftrightarrow \cot {x \over 2} = \cot ( - {\pi \over 3}) \Leftrightarrow {x \over 2} = - {\pi \over 3} + k\pi \cr 
& \Leftrightarrow x = - {{2\pi } \over 3} + k2\pi ;k \in Z \cr} \)

d) \(\tan ({\pi  \over {12}} + 12x) =  - \sqrt 3 \) 

\(\eqalign{
& \tan ({\pi \over {12}} + 12x) = - \sqrt 3 \Leftrightarrow \tan ({\pi \over {12}} + 12\pi ) = \tan ({{ - \pi } \over 3}) \cr 
& \Leftrightarrow {\pi \over {12}} + 12 = {{ - \pi } \over 3} + k\pi \Leftrightarrow x = - {{5\pi } \over {144}} + k{\pi \over {12}},k \in Z \cr} \)