Question

10cos²x-5sinxcosx+3sin²x=4

Answer 1

đáp án có phải vô nghiệm không ạ

 

Answer 2

có nghiệm

Answer 3

\(10cos^2x-5sinx.cosx+3sin^2x=4\)

\(\Leftrightarrow20cos^2x-10-10sinx.cosx+6sin^2x-3+5=0\)

\(\Leftrightarrow7cos2x-5sin2x=-5\)

\(\Leftrightarrow\sqrt{74}\left(\dfrac{7}{\sqrt{74}}cos2x-\dfrac{5}{\sqrt{74}}sin2x\right)=-5\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{7}{\sqrt{74}}\right)=-\dfrac{5}{\sqrt{74}}\)

\(\Leftrightarrow2x+arccos\dfrac{7}{\sqrt{74}}=\pm arccos\left(-\dfrac{5}{\sqrt{74}}\right)+k2\pi\)

\(\Leftrightarrow x=-\dfrac{1}{2}arccos\dfrac{7}{\sqrt{74}}\pm\dfrac{1}{2}arccos\left(-\dfrac{5}{\sqrt{74}}\right)+k\pi\)