Pt \(\Leftrightarrow\dfrac{3}{5}sin\left(x+1\right)+\dfrac{4}{5}cos\left(x+1\right)=1\)
Đặt \(cos\alpha=\dfrac{3}{5}\Rightarrow sin\alpha=\dfrac{4}{5}\) ( vì \(cos^2\alpha+sin^2\alpha=1\))
Pt tt: \(sin\left(x+1\right).cos\alpha+cos\left(x+1\right).sin\alpha=1\)
\(\Leftrightarrow sin\left(x+1+\alpha\right)=1\)
\(\Leftrightarrow x+1+\alpha=\dfrac{\pi}{2}+k2\pi\) (\(k\in Z\))
\(\Leftrightarrow x=\dfrac{\pi}{2}-\alpha-1+k2\pi\) (\(k\in Z\))
`3sin(x+1)+4cos(x+1)=5`
`<=> 3/5 sin(x+1) + 4/5 cos (x+1)=1`
Vì `(3/5)^2 + (4/5)^2 = 1` nên ta có:
Đặt \(\left\{{}\begin{matrix}sinα=\dfrac{3}{5}\\cosα=\dfrac{4}{5}\end{matrix}\right.\)
`sin α . sin(x+1)+cosa . cos(x+1)=1`
`<=> cos(α - x-1)=1`
`<=> α -x-1=k2π`
`<=> x=α-1+k2α (k \in ZZ)`
