Cho cosx = 2/ căn bậc hai 5, 0 < x < pi/2. Tính các giá trị lượng giác của góc x

Ta có: 0 < x < \(\frac{\pi }{2} \Rightarrow \left\{ {\begin{array}{*{20}{c}}{\sin x > 0}\\{\cos x > 0}\end{array}} \right.\)

+) \({\cos ^2}x + {\sin ^2}x = 1 \Leftrightarrow {\frac{2}{{\sqrt 5 }}^2} + {\sin ^2}x = 1\)

\( \Leftrightarrow {\sin ^2}x = \frac{1}{5} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = \frac{1}{{\sqrt 5 }}\left( {TM} \right)}\\{\sin x = - \frac{1}{{\sqrt 5 }}\left( L \right)}\end{array}} \right.\)

\( + )1 + {\cos ^2}x = \frac{1}{{{{\cos }^2}x}} \Leftrightarrow 1 + {\cos ^2}x = \frac{1}{{{{\left( {\frac{2}{{\sqrt 5 }}} \right)}^2}}}\)

\( \Leftrightarrow {\tan ^2}x = \frac{1}{4} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \frac{1}{2}(TM)}\\{{\mathop{\rm t}\nolimits} = - \frac{1}{2}(L)}\end{array}} \right.\).