Question

giải pt sau : 1) \(sin^23x+cos^2x=1\)

                   2)\(sin^2\left(2x+\dfrac{\pi}{5}\right)=cos^2x\)

                   3) \(2sin^2x-sin3x+sin5x=0\)

                   4)\(4sin^3x=sin3x+sin5x\)

                   5)\(sin\left(\pi sin\left(x-\dfrac{\pi}{3}\right)\right)=1\)

Answer 1

1: \(\Leftrightarrow\sin^23x=\sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin3x=\sin x\\\sin3x=\sin\left(-x\right)\end{matrix}\right.\)

TH1: sin 3x=sin x

\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\Pi\\3x=\Pi-x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\Pi\\x=\dfrac{1}{4}\Pi+\dfrac{1}{2}k\Pi\end{matrix}\right.\)

5: \(\Leftrightarrow\Pi\cdot\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{\Pi}{2}+k\Pi\)

\(\Leftrightarrow\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{1}{2}+k\)

Để pt có nghiệm thì k+1/2>=-1 và k+1/2<=1

=>-3/2<=k<=1/2

Khi đó, pt sẽ có nghiệm là:

\(\left\{{}\begin{matrix}x-\dfrac{\Pi}{3}=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\\x-\dfrac{\Pi}{3}=\Pi-arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi+\dfrac{\Pi}{3}\\x=\dfrac{4}{3}\Pi+arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)