Question

giải pt : 1) \(\dfrac{1}{sin^2x}=cotx+3\)

            2)\(sin^4x+cos^4x+cos\left(x-\dfrac{\pi}{4}\right)sin\left(3x-\dfrac{\pi}{4}\right)=0\)

 

Answer 1

1) ĐK : \(x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

 \(...\Leftrightarrow1+cot^2x=cotx+3\)  \(\Leftrightarrow cot^2x-cotx-2=0\)  

\(\Leftrightarrow\left(cotx-2\right)\left(cotx+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}cotx=2\\cotx=-1\end{matrix}\right.\)

Đến đây ; bn làm tiếp

2) \(...\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[sin\left(4x-\dfrac{\pi}{2}\right)+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[-cos4x+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[2sin^22x-1+sin2x\right]=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(sin^22x+sin2x+1\right)=0\) \(\Leftrightarrow PTVN\)