Đáp án:
`a, S = {± (3π)/4 + k2π | k \in ZZ}`
`b, S= {π/3+kπ; π/4+kπ | k \ in ZZ}`
Giải thích các bước giải:
`a, (2cos x +\sqrt{2})(cos x-2)=0`
`<=>`\(\left[ \begin{array}{l}2cos x +\sqrt{2} =0 \\cos x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}cos x =-\frac{\sqrt{2}}{2} \\cos x=2 \ (vô \ nghiệm ) \end{array} \right.\)
`<=> cos x = cos (3π)/4`
`<=> x = ± (3π)/4 + k2π \ (k \ in ZZ)`
Vậy `S = {± (3π)/4 + k2π | k \in ZZ}`
`b, (tan x -\sqrt{3})(1-tan x)=0`
`<=>`\(\left[ \begin{array}{l}tan x -\sqrt{3}=0\\1-tan x =0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}tan x =\sqrt{3}\\tan x =1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}tan x =tan \fracπ3\\tan x =tan\fracπ4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l} x =\fracπ3+kπ\\ x =\fracπ4 +kπ\end{array} \right.\) `(k \in ZZ)`
Vậy `S= {π/3+kπ; π/4+kπ | k \ in ZZ}`
