Ta có:[imath]63 + 60cos4x + 5cos8x = 8sin^5 2x[/imath]
[imath]\iff 63+ 60(1 - 2sin^2 2x) + 5(2cos^2 4x - 1) = 8sin^5 2x[/imath]
[imath]\iff 63 + 60 - 120sin^2 2x + 5[2(1-2sin2x)^2 - 1] = 8sin^5 2x[/imath]
[imath]\iff 123 - 120sin^2 2x + 5(1 - 8sin2x + 8sin^2 2x) - 8sin^5 2x = 0[/imath]
[imath]\iff 128 - 80sin^2 2x - 40sin2x - 8sin^5 2x = 0[/imath]
[imath]\iff (sin2x - 1)(8sin^4 2x + 8sin^3 2x + 8sin^2 2x + 88sin2x + 128)=0[/imath]
[imath]\iff (sin2x - 1)(sin^4 2x + sin^3 2x + sin^2 2x + 11sin2x + 16)=0[/imath]
[imath]\iff \left[\begin{matrix} sin2x - 1 = 0\\ sin^4 2x + sin^3 2x + sin^2 2x + 11sin2x + 16=0\end{matrix}\right.[/imath]
Với [imath]sin2x = 1[/imath]
[imath]\iff 2x = \dfrac{\pi}{2} + 2k\pi[/imath]
[imath]\iff x = \dfrac{\pi}{4} + k\pi(k \in \mathbb{Z})[/imath]
Do
[imath]0 \leq sin^4 2x \leq 1[/imath]
[imath]0 \leq sin^2 2x \leq 1[/imath]
[imath]-1 \leq sin^3 2x \leq 1[/imath]
[imath]-11 \leq 11sin2x \leq 11[/imath]
[imath]\to 4 \leq sin^4 2x + sin^3 2x + sin^2 2x + 11sin2x + 16 \leq 30[/imath]
[imath]\to sin^4 2x + sin^3 2x + sin^2 2x + 11sin2x + 16 \neq 0[/imath]
[imath]\to \text{phương trình vô nghiệm}[/imath]
Vậy phương trình đã cho có tập nghiệm
[imath]S =[/imath] {[imath]\dfrac{\pi}{4} + k\pi|k \in \mathbb{Z}[/imath]}
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
