Từ phương trình ban đầu ta có : \(2\cos5x\sin x=\sqrt{3}\sin^2x+\sin x\cos x\)
\(\Leftrightarrow\begin{cases}\sin x=0\\2\cos5x=\sqrt{3}\sin x+\cos x\end{cases}\)
+) \(\sin x=0\Leftrightarrow x=k\pi\)
+)\(2\cos5x=\sqrt{3}\sin x+\cos x\Leftrightarrow\cos5x=\cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{18}+\frac{k\pi}{3}\end{cases}\)
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
x=π/2+kπ
x=π/18+kπ/9
x = π/2 + kπ
x = π/18 + kπ/9
x=π/2+kπ
x=π/18+kπ/9
1/2(cos4x+cos2x)+1/2(cos8x−cos4x)+1/2(cos10x−cos2x)=0⇔1/2(cos4x+cos2x)+1/2(cos8x−cos4x)+1/2(cos10x−cos2x)=0
⇔1/2cos4x+1/2cos2x+1/2cos8x−1/2cos4x)+1/2cos10x−1/2cos2x=
Đúng(0)
x = π/18 + kπ/9 , π/2 + kπ , kϵz
π/2+kπ
π/18+kπ9
x=π/2+kπ
x=π/18 +kπ/9
cosx.cos3x−sin2x.sin6x−sin4x.sin6x=0.
Hướng dẫn giải:
\(cos x . cos 3 x - sin 2 x . sin 6 x - sin 4 x . sin 6 x = 0 \left(\right. 1 \left.\right)\)
(Dạng phương trình biến đổi tích thành tổng)
\(\left(\right. 1 \left.\right) \Leftrightarrow \frac{1}{2} \left(\right. cos 4 x + cos 2 x \left.\right) - \frac{1}{2} \left(\right. cos 4 x - cos 8 x \left.\right) - \frac{1}{2} \left(\right. cos 2 x - cos 10 x \left.\right) = 0\)
\(\Leftrightarrow cos 10 x + cos 8 x = 0\)\(\Leftrightarrow 2 cos 9 x . cos x = 0 \Leftrightarrow \left[\right. cos 9 x = 0 \\ cos x = 0\)
\(\Leftrightarrow \left[\right. x = \frac{\pi}{18} + \frac{k \pi}{9} \\ x = \frac{\pi}{2} + k \pi\)\(\left(\right. k \in \mathbb{Z} \left.\right)\)
Vậy các nghiệm của phương trình là \(x = \frac{\pi}{18} + \frac{k \pi}{9}\) và \(x = \frac{\pi}{2} + k \pi\) \(\left(\right. k \in \mathbb{Z} \left.\right)\).