Đáp án:
$a,$
\(\left[ \begin{array}{l}-x=-30^0+k2\pi\\x=-10^0+\frac{k2\pi}{3}\end{array} \right.\) $k∈Z$
$b,$
\(\left[ \begin{array}{l}x=\frac{15\pi}{8}+k2\pi\\x=\frac{11\pi}{8}+k2\pi\end{array} \right.\) $k∈Z$
Giải thích các bước giải:
$a, cos(x^{}+30^0)=cos2x$
⇔\(\left[ \begin{array}{l}x+30^0=2x+k2\pi\\x+30^0=-2x+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}-x=-30^0+k2\pi\\x=-10^0+\frac{k2\pi}{3}\end{array} \right.\) $k∈Z$
$b, cos(x^{}-\frac{13\pi}{8})=$ $\frac{\sqrt2}{2}$
⇔$cos(x^{}-\frac{13\pi}{8})=cos\frac{\pi}{4}$
⇔\(\left[ \begin{array}{l}x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\\x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{15\pi}{8}+k2\pi\\x=\frac{11\pi}{8}+k2\pi\end{array} \right.\) $k∈Z$
