Blog Toán Phổ Thông
Bài giảng và tài liệu toán phổ thông file word
a) sin^2 x + sin^2 2x + sin^2 3x =2
b) 2cos^2 2x + cos2x = 4sin^2 2x cos^2 x
c) 2cos6x + tan3x = 4/5
d) (1 – tanx)(1 + sin2x) = 1 + tanx
Có 1 trả lời
\[\begin{array}{l}a)\,\,{\sin ^2}x + {\sin ^2}2x + {\sin ^2}3x = 2\\\Leftrightarrow \frac{{1 – \cos 2x}}{2} + \frac{{1 – \cos 4x}}{2} + \frac{{1 – \cos 6x}}{2} = 2\\\Leftrightarrow \cos 2x + \cos 4x + \cos 6x =-1\\\Leftrightarrow \cos 2x + 2\cos 2x – 1 + 4{\cos ^3}2x – 3\cos 2x =-1\\\Leftrightarrow \cos 2x = 0\end{array}\]
\[\begin{array}{l}a)\,\,{\sin ^2}x + {\sin ^2}2x + {\sin ^2}3x = 2\\\Leftrightarrow \frac{{1 – \cos 2x}}{2} + \frac{{1 – \cos 4x}}{2} + \frac{{1 – \cos 6x}}{2} = 2\\\Leftrightarrow \cos 2x + \cos 4x + \cos 6x =-1\\\Leftrightarrow \cos 2x + 2\cos 2x – 1 + 4{\cos ^3}2x – 3\cos 2x =-1\\\Leftrightarrow \cos 2x = 0\end{array}\]