Giải phương trình:
LG a
\({2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)
Lời giải chi tiết:
Ta có: \(\,{2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)
\(\Leftrightarrow {2^{1 - {{\cos }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)
\( \Leftrightarrow \frac{2}{{{2^{{{\cos }^2}x}}}} + {4.2^{{{\cos }^2}x}} = 6\)
Đặt \(t = {2^{{{\cos }^2}x}}\)
Vì \(0 \le {\cos ^2}x \le 1 \) \(\Rightarrow {2^0} \le {2^{{{\cos }^2}x}} \le {2^1}\) \( \Rightarrow 1 \le t \le 2\)
Ta có:
\({2 \over t} + 4t = 6 \)\(\Leftrightarrow 4{t^2} - 6t + 2 = 0\)
\(\Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = {1 \over 2}\,\,\left( \text{loại} \right) \hfill \cr} \right.\)
\( \Leftrightarrow {2^{{{\cos }^2}x}} = 1 \Leftrightarrow {\cos ^2}x = 0\)
\(\Leftrightarrow \cos x = 0 \Leftrightarrow x = {\pi \over 2} + k\pi ,\,k \in \mathbb Z\)
LG b
\({4^{3 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = {4^{{1 \over 2}}}\)
Lời giải chi tiết:
\({4^{3 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = {4^{{1 \over 2}}}\)
\( \Leftrightarrow {4}{.4^{2 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = 2\)
\( \Leftrightarrow {4.4^{2\left( {1 + \cos 2x} \right)}} - {7.4^{1 + \cos 2x}} = 2\)
Đặt \(t = {4^{1 + \cos 2x}}\,\left( {t > 0} \right)\)
Ta có:
\(\eqalign{
& 4{t^2} - 7t - 2 = 0 \Leftrightarrow \left[ \matrix{
t = 2 \hfill \cr
t = - {1 \over 4}\,\left( \text {loại} \right) \hfill \cr} \right. \cr
& \Leftrightarrow {2^{2 + 2\cos 2x}} = 2 \Leftrightarrow 2 + 2\cos 2x = 1 \cr
& \Leftrightarrow \cos 2x = - {1 \over 2} = \cos {{2\pi } \over 3} \cr
& \Leftrightarrow 2x = \pm {2\pi \over 3} + k2\pi ,\,k \in \mathbb Z \cr} \)
\( \Leftrightarrow x = \pm \frac{\pi }{3} + k\pi ,k \in Z\)
