Đáp án: $ x=\pm\arccos(\dfrac{5\pm4\sqrt7}{29})+k2\pi, k\in Z$
Giải thích các bước giải:
Ta có:
$2\sin x+5\cos x=1$
$\to 2\sin x=1-5\cos x$
$\to (2\sin x)^2=(1-5\cos x)^2$
$\to 4\sin^2x=1-10\cos x+25\cos^2x$
$\to 4(1-\cos^2x)=1-10\cos x+25\cos^2x$
$\to 4-4\cos^2x=1-10\cos x+25\cos^2x$
$\to 29\cos^2x-10\cos x-3=0$
$\to \cos x=\dfrac{5\pm4\sqrt7}{29}$
$\to x=\pm\arccos(\dfrac{5\pm4\sqrt7}{29})+k2\pi, k\in Z$
