$\text{ Ta có sin x + cos x = 5/4 }$
$\text{ => sin x = 5/4 - cos x}$
$\text{ mà $sin^2x + cos^2x = 1$}$
<=> $(5/4 - cos x)^2 + cos^2x = 1$<=> $25/16 - 10cos x/4+cos^2x + cos^2x = 1$
<=> $ 2cos^2x - 10cos x/4+25/16 = 1$
Đặt $cos x = a ( -1 ≤ a ≤ 1)$
=> $ 2a^2 - 10a/4+25/16 = 1$
=> $ 2a^2 - 10a/4+9/16 = 0$
=> a = $\dfrac{5-\sqrt[]{7}}{8}$ (TM)
hoặc a = $\dfrac{5+\sqrt[]{7}}{8}$ (TM)
=> +) cos x = $\dfrac{5-\sqrt[]{7}}{8}$
=> sin x = $\dfrac{5+\sqrt[]{7}}{8}$
hoặc cos x = $\dfrac{5+\sqrt[]{7}}{8}$
sin x = $\dfrac{5-\sqrt[]{7}}{8}$
=> sin x.cos x = $\dfrac{5+\sqrt[]{7}}{8}$.$\dfrac{5-\sqrt[]{7}}{8}$ = $\dfrac{9}{32}$
