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Bài 77. Giải phương trình:
\(a)\,{2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\,;\)
\(b)\,{4^{3 + 2\cos 2x}} – {7.4^{1 + \cos 2x}} = {4^{{1 \over 2}}}\)
a) Ta có: \(\,{2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\, \Leftrightarrow {2^{1 – {{\cos }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)
Đặt \(t = {2^{{{\cos }^2}x}}\,\left( {1 \le t \le 2} \right)\)
Ta có:
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\({2 \over t} + 4t = 6 \Leftrightarrow 4{t^2} – 6t + 2 = 0 \Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = {1 \over 2}\,\,\left( \text{loại} \right) \hfill \cr} \right.\)
\( \Leftrightarrow {2^{{{\cos }^2}x}} = 1 \Leftrightarrow \cos x = 0 \Leftrightarrow x = {\pi \over 2} + k\pi ,\,k \in \mathbb Z\)
b) Đặt \(t = {4^{t + \cos 2x}}\,\left( {t > 0} \right)\)
Ta có: \({4.4^{2\left( {1 + \cos 2x} \right)}} – {7.4^{1 + \cos 2x}} = 2\)
\(\eqalign{
& \Leftrightarrow 4{t^2} – 7t – 2 = 0 \Leftrightarrow \left[ \matrix{
t = 2 \hfill \cr
t = – {1 \over 4}\,\left( \text {loại} \right) \hfill \cr} \right. \cr
& \Leftrightarrow {2^{2 + 2\cos 2x}} = 2 \Leftrightarrow 2 + 2\cos 2x = 1 \cr
& \Leftrightarrow \cos 2x = – {1 \over 2} = \cos {{2\pi } \over 3} \cr
& \Leftrightarrow x = \pm {2\pi \over 3} + k\pi ,\,k \in \mathbb Z \cr} \)