Đáp án:
a, \(\left[ \begin{array}{l}x=\frac{\pi}{4}-k\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{array} \right.\) (k∈Z)
b, \(\left[ \begin{array}{l}x=\frac{\pi}{4}-k2\pi\\x=\frac{\pi}{4}+\frac{2k\pi}{3}\end{array} \right.\) (k∈Z)
Giải thích các bước giải:
a, cosx - sin3x = 0
⇔ cosx = sin3x
⇔ cosx = cos(3x-$\frac{\pi}{2}$)
⇔ \(\left[ \begin{array}{l}x=3x-\frac{\pi}{2}+k2\pi\\x=-3x+\frac{\pi}{2}+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}-2x=-\frac{\pi}{2}+k2\pi\\4x=\frac{\pi}{2}+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}x=\frac{\pi}{4}-k\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{array} \right.\) (k∈Z)
b, sinx + cosx = $\sqrt[]{2}$sin2x
⇔ $\frac{1}{\sqrt[]{2}}$sinx + $\frac{1}{\sqrt[]{2}}$cosx = sin2x
⇔ sin(x+$\frac{\pi}{4}$) = sin2x
⇔ \(\left[ \begin{array}{l}x+\frac{\pi}{4}=2x+k2\pi\\x+\frac{\pi}{4}=\pi -2x+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}-x=-\frac{\pi}{4}+k2\pi\\3x=\frac{3\pi}{4}+k2\pi\end{array} \right.\) (k∈Z)
⇔ \(\left[ \begin{array}{l}x=\frac{\pi}{4}-k2\pi\\x=\frac{\pi}{4}+\frac{2k\pi}{3}\end{array} \right.\) (k∈Z)
