trig inequality qn (1 Viewer)

[Masaken]

i've done part (a), answer being but i don't know how to use this answer for part b? how would i start this qn? thanks in advance

 

[Masaken]

quick correction is that the qn is less than or equal to -1, not 1, oops (the answer is the same though, i just copied it out wrong)

 

[gamja]

just looking quickly - i can tell ur meant to sub in x=cosx+1 into first equation, which leads to the same solution of 1<cosx+1<=3,

therefore 0<cosx<=2

0<x<=pi/2

 

[Masaken]

just looking quickly - i can tell ur meant to sub in x=cosx+1 into first equation, which leads to the same solution of 1<cosx+1<=3,

therefore 0<cosx<=2

0<x<=pi/2

thank you, i get most of it up to the bit where 2 becomes pi/2? how would you get that?? (sorry if the answer's obvious, it's just not clicking)
update: nevermind, i get it thanks!!