Question

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Answer 1

Điều kiện −1 ≤ sinx; cosx ≤ 1.

Ta có: \(\sin x + \cos x = \frac{1}{2} \Leftrightarrow \cos x = \frac{1}{2} - \sin x\)

Mặt khác: sin²x + cos²x = 1 \( \Leftrightarrow {\sin ^2}x + {\left( {\frac{1}{2} - \sin x} \right)^2} = 1\)

\( \Leftrightarrow {\sin ^2}x + {\sin ^2}x - \sin x + \frac{1}{4} = 1\)

\( \Leftrightarrow 2{\sin ^2}x - \sin x - \frac{3}{4} = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{{1 + \sqrt 7 }}{4}\\\sin = \frac{{1 - \sqrt 7 }}{4}\end{array} \right.\)

Ta có:

• \(\sin x = \frac{{1 + \sqrt 7 }}{4} \Rightarrow \cos x = \frac{{1 - \sqrt 7 }}{4}\) (TMĐK)

• \(\sin x = \frac{{1 - \sqrt 7 }}{4} \Rightarrow \cos x = \frac{{1 + \sqrt 7 }}{4}\)(TMĐK)

Vậy \(\sin x = \frac{{1 + \sqrt 7 }}{4};\,\,\cos x = \frac{{1 - \sqrt 7 }}{4}\) hoặc \(\sin x = \frac{{1 - \sqrt 7 }}{4};\,\,\cos x = \frac{{1 + \sqrt 7 }}{4}\)

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Answer 2