Question

Sin6x * cos2x< sin5x * co3x, помогите пожалуйста, хотя направление дайте как решать, очень нужно!!!!!!!

Answer 1

$\frac{1}{2} (sin(6x-2x)+sin(6x+2x))\ \textless \ \frac{1}{2} (sin(5x-3x)+sin(5x+3x))$
sin4x+sin8x<sin2x+sin8x
sin4x-sin2x<0
2sin2xcos2x-sin2x<0
sin2x(2cos2x-1)<0
$\begin {cases} sin2x\ \textgreater \ 0 \\ cos2x\ \textless \ \frac{1}{2} \end {cases}$           или           $\begin {cases} sin2x\ \textless \ 0 \\ cos2x\ \textgreater \ \frac{1}{2} \end {cases}$
$\begin {cases} 2\pi k\ \textless \ 2x\ \textless \ \pi + 2\pi k \\ \frac{\pi}{3}+2 \pi k \ \textless \ 2x\ \textless \ \frac{\pi}{3}+2 \pi k \end {cases}$       или       $\begin {cases} -\pi + 2\pi k\ \textless \ 2x\ \textless \ 2\pi k \\ -\frac{\pi}{3}+2 \pi k \ \textless \ 2x\ \textless \ \frac{\pi}{3}+2 \pi k \end {cases}$
$\frac{\pi}{3}+2\pi k\ \textless \ 2x\ \textless \ \pi +2 \pi k$       или       $-\frac{\pi}{3}+2\pi k\ \textless \ 2x\ \textless \ 2 \pi k$
$\frac{\pi}{6}+\pi k\ \textless \ x\ \textless \ \frac{\pi}{2}+ \pi k$      или      $- \frac{\pi}{6}+\pi k\ \textless \ x\ \textless \ \pi k$
Ответ: $( -\frac{\pi}{6}+\pi k; \pi k ) \cup ( \frac{\pi}{6}+\pi k; \frac{\pi}{2}+ \pi k ),\ k \in Z$

Answer 2

1/2sin(6x-2x)+1/2sin(6x+2x)<1/2sin(5x-3x)+1/2sin(5x+3x)
1/2sin4x+1/2sin8x<1/2sin2x+1/2sin8x
1/2sin4x-1/2sin2x<0
1/2*2sin2xcos2x-1/2sin2x<0
1/2sin2x(2cos2x-1)<0
1){sin2x>0⇒2πk<2x<π+2πk⇒πk<x<π/2+πk,k∈z
{cos2x<1/2⇒π/3+2πk<2x<5π/3+2πk⇒π/6+πk<x<5π/6+πk,k∈z
x∈(π/6+πk;π/2+πk,k)
2){sin2x<0⇒π+2πk<2x<2π+2πk⇒π/2+πk<x<π+πk,k∈z
{cos2x>1/2⇒-π/3+2πk<2x<π/3+2πk⇒-π/6+πk<x<π/6+πk,k∈z
x∈(-π/6+πk;πk,k∈Z)
Ответ x∈(-π/6+πk;πk,k∈Z) U (π/6+πk;π/2+πk,k)