Giải phương trình lượng giác

Question

a) $sin \left ( x + \frac{\pi }{6}\right ) = cos2x$;

b) $sinx-cosx = 1$.

score 0 · Answer 1

b) $\sin x-\cos x=\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)$

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

$\sin x-\cos x=1\Leftrightarrow \sqrt{2}\sin\left( x-\dfrac{\pi}{4}\right) =1$

$\Leftrightarrow \sin \left( x-\dfrac{\pi}{4}\right) =\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=\sin\dfrac{\pi}{4}$

$\Leftrightarrow\left[\begin{matrix} x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ \\ x-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[\begin{matrix}x= \dfrac{\pi}{2}+k2\pi \qquad (k\in Z) \\ \\ x=\pi +k2\pi\qquad (k\in Z)\end{matrix}\right.$

score 0 · Answer 2

Bình chọn tăng 0 Bình chọn giảm

a) $\sin\left( x+\dfrac{\pi}{6}\right) =\cos 2x\Leftrightarrow \sin\left( x+\dfrac{\pi}{6}\right) =\sin\left( \dfrac{\pi}{2}-2x\right)$

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

$\Leftrightarrow\left[\begin{matrix}x=\dfrac{\pi}{9}+k\dfrac{2\pi}{3}\qquad (k\in Z) \\ x=-\dfrac{\pi}{3}+k2\pi \qquad (k\in Z)\end{matrix}\right.$