Question

Giaỉ các phương trình lượng giác sau:

1. sin(sinx)=0

2. sin(cosx)=0

3. \(\sqrt{3}\sin-\cos x=2cos3x\)

4. \(\sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)

5. \(4\cos\left(3\pi-2x\right)=\sqrt{2}\)

Answer 1

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

Answer 2

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

Answer 3

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

Answer 4

4.

\(sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2x-\dfrac{\pi}{2}+k2\pi\left(vn\right)\\2x=\dfrac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{3\pi}{8}+\dfrac{k\pi}{2}\)

5.

\(\Leftrightarrow cos\left(3\pi-2x\right)=\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow cos2x=-\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(-\dfrac{\sqrt{2}}{4}\right)+k\pi\)