Question

12t-2<-5t+36
8x-8<=-72

2cos^2(x)-sqrt(3)cos(x)=0, 0^{\circ}< x< 360^{\circ}

ayuda

Answer 1

Respuesta:

12t-2<-5t+36

$12t-2+2<-5t+36+2\\\\12t<-5t+38\\\\12t+5t<-5t+38+5t\\\\17t<38\\\\\frac{17t}{17}<\frac{38}{17}\\\\t<\frac{38}{17}$

8x-8<=-72

$8x-8+8\le \:-72+8\\\\8x\le \:-64\\\\\frac{8x}{8}\le \frac{-64}{8}\\\\x\le \:-8$

2cos^2(x)-sqrt(3)cos(x)=0, 0^{\circ}< x< 360^{\circ}

$\cos \left(x\right)=\frac{\sqrt{3}}{2},\:\cos \left(x\right)=0\\\\x=30^{\circ \:},\:x=330^{\circ \:},\:x=90^{\circ \:},\:x=270^{\circ \:}$

Answer 2

Respuesta:

Explicación paso a paso: