Question

Giải phương trình : cosx + 5sinx/2 - 3 = 0

Answer 1

cosx + \(\dfrac{5}{2}\)sinx = 3

vì 1² + 2,5² < 3² nên pt vô nghiệm

Answer 2

\(cosx+5sin\dfrac{x}{2}-3\)=0

<=> 1-2\(sin^2\dfrac{x}{2}\)+\(5sin\dfrac{x}{2}\)-3=0

<=>2\(sin^2\dfrac{x}{2}-5sin\dfrac{x}{2}\)+2=0

<=>2si\(n^2\dfrac{x}{2}-4sin\dfrac{x}{2}-sin\dfrac{x}{2}+2\)=0

<=>(\(sin\dfrac{x}{2}-2\))(\(2sin\dfrac{x}{2}-1\))=0

\(\left[{}\begin{matrix}sin\dfrac{x}{2}=2\left(lọại\right)\\sin\dfrac{x}{2}=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

vơi \(sin\dfrac{x}{2}=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{\pi}{6}+k2\pi\\\dfrac{x}{2}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k4\pi\\x=\dfrac{5\pi}{3}+k4\pi\end{matrix}\right.\)