Question

sinx+cosx=1-\(\dfrac{1}{2}\).sin2x

Answer 1

Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Pt trở thành:

\(t=1-\dfrac{t^2-1}{4}\) \(\Leftrightarrow t^2+4t-5=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=1\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)