5* $cos^{2}${x} + 7* $sin^{2}${x} = 6.5 Tìm x

Đáp án:

$\left[\begin{array}{l}  x=  \dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{2\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{-\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{4\pi}{3} +k 2 \pi(k \in \mathbb{Z})\end{array} \right..$

Giải thích các bước giải:

$5\cos^2x+7\sin^2x=6,5\\ \Leftrightarrow 5\cos^2x+5\sin^2x+2\sin^2x=6,5\\ \Leftrightarrow 5(\cos^2x+\sin^2x)+2\sin^2x=6,5\\ \Leftrightarrow 5+2\sin^2x=6,5\\ \Leftrightarrow 2\sin^2x=1,5\\ \Leftrightarrow \sin^2x=\dfrac{3}{4}\\ \Leftrightarrow\left[\begin{array}{l} \sin x= \dfrac{\sqrt{3}}{2} \\\sin x= -\dfrac{\sqrt{3}}{2} \end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} \sin x= \sin \dfrac{\pi}{3} \\\sin x= \sin \dfrac{-\pi}{3} \end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l}  x=  \dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{2\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{-\pi}{3} +k 2 \pi(k \in \mathbb{Z}) \\ x=  \dfrac{4\pi}{3} +k 2 \pi(k \in \mathbb{Z})\end{array} \right..$