Đáp án:
Giải thích các bước giải:
a, cos(x-$\frac{\pi}{4}$)=1
⇔cos(x-$\frac{\pi}{4}$)=cos (0)
⇔x-$\frac{\pi}{4}$=0
⇔x=$\frac{\pi}{4}$+k2π k∈Z
b, sin²x-3sinx+2=0
⇔\(\left[ \begin{array}{l}sinx=2 (loại)\\sinx=1\end{array} \right.\)
⇔sinx=1
⇔x=$\frac{\pi}{2}$ +k2$\pi$ k∈Z
c, √3sinx-cosx=2sin2x
⇔$\frac{\sqrt{3}}{2}$ sinx-$\frac{1}{2}$ cosx=sin2x
⇔cos$\frac{\pi}{6}$ sinx-sin$\frac{\pi}{6}$ cosx=sin2x
⇔sin(x-$\frac{\pi}{6}$ )=sin2x
⇔\(\left[ \begin{array}{l}x-\frac\pi6=2x+k2\pi\\x-\frac\pi6=\pi-2x+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-\pi}6+k2\pi\\x=\frac{7\pi}{18}+\frac{k2\pi}3\end{array} \right.\)
