Giải giúp em câu này với ạ. Em cảm ơn! Cos³x + Cos⁴x = Sin³x + Sin⁴x

Đáp án:

$\left[\begin{array}{l}x =\dfrac{\pi}{4}+k\pi\\ x = \pi + k2\pi\\ x = -\dfrac{\pi}{2}+ k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Giải thích các bước giải:

$\cos^3x +\cos^4x =\sin^3x \sin^4x$

$\to (\cos^3x -\sin^3x) + (\cos^4x -\sin^4x)=0$

$\to (\cos x -\sin x)(\cos^2x +\cos x\sin x+\sin^2x) + (\cos^2x -\sin^2x)(\cos^2x +\sin^2x)=0$

$\to (\cos x -\sin x)(1+\cos x\sin x) + (\cos x -\sin x)(\cos x +\sin x) = 0$

$\to (\cos x -\sin x)(1+\cos x\sin x +\cos x +\sin x)=0$

$\to (\cos x -\sin x)(1+\cos x)(1+\sin x) = 0$

$\to \left[\begin{array}{l}\cos x -\sin x = 0\\\cos x +1 = 0\\\sin x + 1 = 0\end{array}\right.$

$\to \left[\begin{array}{l}\cos\left(x +\dfrac{\pi}{4}\right) = 0\\\cos x = -1\\\sin x = -1\end{array}\right.$

$\to \left[\begin{array}{l}x +\dfrac{\pi}{4}= \dfrac{\pi}{2} + k\pi\\ x = \pi + k2\pi\\ x = -\dfrac{\pi}{2}+ k2\pi\end{array}\right.$

$\to \left[\begin{array}{l}x =\dfrac{\pi}{4}+k\pi\\ x = \pi + k2\pi\\ x = -\dfrac{\pi}{2}+ k2\pi\end{array}\right.\quad (k\in\Bbb Z)$