\(\left(\sin x+\cos x\right)^2=\dfrac{9}{16}\)
nên \(1+2\cdot\sin x\cdot\cos x=\dfrac{9}{16}\)
\(\Leftrightarrow\sin x\cdot\cos x=-\dfrac{7}{32}\)
\(\left|\sin x-\cos x\right|=\sqrt{\left(\sin x+\cos x\right)^2-4\cdot\sin x\cdot\cos x}\)
\(=\sqrt{\dfrac{9}{16}-4\cdot\dfrac{-7}{32}}=\dfrac{\sqrt{23}}{4}\)
nên \(\left[{}\begin{matrix}\sin x-\cos x=\dfrac{\sqrt[4]{23}}{2}\\\sin x-\cos x=-\dfrac{\sqrt[4]{23}}{2}\end{matrix}\right.\)
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