Giải phương trình a)cos(x+30°)=cos2x b)cos(x-13pi/8)=✓2/2 10/08/2021 Bởi Reagan Giải phương trình a)cos(x+30°)=cos2x b)cos(x-13pi/8)=✓2/2
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: $a, cos(x^{}+30^0)=cos2x$ ⇔\(\left[ \begin{array}{l}x+30^0=2x+k2\pi\\x+30^0=-2x+k2\pi\end{array} \right.\) ⇔\(\left[ \begin{array}{l}-x=-30^0+k2\pi\\x=-10^0+\frac{k2\pi}{3}\end{array} \right.\) $k∈Z$ $b, cos(x^{}-\frac{13\pi}{8})=$ $\frac{\sqrt2}{2}$ ⇔$cos(x^{}-\frac{13\pi}{8})=cos\frac{\pi}{4}$ ⇔\(\left[ \begin{array}{l}x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\\x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{15\pi}{8}+k2\pi\\x=\frac{11\pi}{8}+k2\pi\end{array} \right.\) $k∈Z$ Bình luận
a) $cos(x+30^o)=cos2x$ $↔ \left[ \begin{array}{l}x+30^o=2x+k2\pi\\x+30^o=-2x+k2\pi\end{array} \right.$ $↔ \left[ \begin{array}{l}x=30^o-k2\pi\\x=-10^o+\dfrac{k2\pi}{3}\end{array} \right.$ b) $cos(x-\dfrac{13\pi}{8})=\dfrac{\sqrt[]{2}}{2}$ $↔ cos(x-\dfrac{13\pi}{8})=cos\dfrac{\pi}{4}$ $↔ \left[ \begin{array}{l}x-\dfrac{13\pi}{8}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{13\pi}{8}=-\dfrac{\pi}{4}+k2\pi\end{array} \right.$ $↔ \left[ \begin{array}{l}x=\dfrac{15\pi}{8}+k2\pi\\x=\dfrac{11\pi}{8}+k2\pi\end{array} \right.$ Bình luận
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:
$a, cos(x^{}+30^0)=cos2x$
⇔\(\left[ \begin{array}{l}x+30^0=2x+k2\pi\\x+30^0=-2x+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}-x=-30^0+k2\pi\\x=-10^0+\frac{k2\pi}{3}\end{array} \right.\) $k∈Z$
$b, cos(x^{}-\frac{13\pi}{8})=$ $\frac{\sqrt2}{2}$
⇔$cos(x^{}-\frac{13\pi}{8})=cos\frac{\pi}{4}$
⇔\(\left[ \begin{array}{l}x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\\x-\frac{13\pi}{8}=\frac{\pi}{4}+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{15\pi}{8}+k2\pi\\x=\frac{11\pi}{8}+k2\pi\end{array} \right.\) $k∈Z$
a) $cos(x+30^o)=cos2x$
$↔ \left[ \begin{array}{l}x+30^o=2x+k2\pi\\x+30^o=-2x+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=30^o-k2\pi\\x=-10^o+\dfrac{k2\pi}{3}\end{array} \right.$
b) $cos(x-\dfrac{13\pi}{8})=\dfrac{\sqrt[]{2}}{2}$
$↔ cos(x-\dfrac{13\pi}{8})=cos\dfrac{\pi}{4}$
$↔ \left[ \begin{array}{l}x-\dfrac{13\pi}{8}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{13\pi}{8}=-\dfrac{\pi}{4}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{15\pi}{8}+k2\pi\\x=\dfrac{11\pi}{8}+k2\pi\end{array} \right.$