Question

giải pt

a) \(\sqrt{3}sinx+cosx=2\)

b) \(sin\left(\frac{\pi}{4}-2x\right)+sin\left(\frac{\pi}{4}+x\right)=\frac{\sqrt{2}}{2}\)

Answer 1

\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k2\pi\)

b.

\(\sqrt{2}sin\left(\frac{\pi}{4}-2x\right)+\sqrt{2}sin\left(\frac{\pi}{4}+x\right)=1\)

\(\Leftrightarrow cos2x-sin2x+sinx+cosx=1\)

\(\Leftrightarrow1-2sin^2x-2sinx.cosx+sinx+cosx=1\)

\(\Leftrightarrow-2sinx\left(sinx+cosx\right)+sinx+cosx=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow x=...\)