Question

\(\sin4x\sin5x+\sin4x\sin3x-\sin2x\sin x=0\)

Answer 1

\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)

\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)

\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)

\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)

\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{k\pi}{3}\\x=\frac{k\pi}{5}\end{matrix}\right.\)

Answer 2

\(sin ⁡ 5 x - sin ⁡ 4 x + sin ⁡ 3 x \left(\right. 1 \left.\right)\)

\(\Leftrightarrow \left(\right. sin ⁡ 3 x + sin ⁡ 5 x \left.\right) - sin ⁡ 4 x = 0\)

\(\Leftrightarrow 2 sin ⁡ 4 x . cos ⁡ x - sin ⁡ 4 x = 0\)

\(\Leftrightarrow sin ⁡ 4 x \left(\right. 2 cos ⁡ x - 1 \left.\right) = 0\)

\(\Leftrightarrow \left[\right. sin ⁡ 4 x = 0 \\ 2 cos ⁡ x - 1 = 0 \Leftrightarrow \left[\right. x = \frac{k \pi}{4} \\ x = \frac{\pi}{3} + k 2 \pi \\ x = - \frac{\pi}{3} + k 2 \pi\)\(\left(\right. k \in \mathbb{Z} \left.\right)\)

Vậy các nghiệm của phương trình là \(x = \frac{k \pi}{4}\), \(x = \frac{\pi}{3} + k 2 \pi\) và  \(x = - \frac{\pi}{3} + k 2 \pi\) \(\left(\right. k \in \mathbb{Z} \left.\right)\).