November 9th, 2016, 01:24 PM  #1 
Newbie Joined: Nov 2016 From: Estonia Posts: 2 Thanks: 0  Please help me :)
a) 3sinx = 2,1 b) 2 + sinx = 0 c) cos³x = 0,512 d) cos²x  3sin²x = 0 e) cos²x + 3cosx = 0 
November 9th, 2016, 01:57 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 16,397 Thanks: 1177 
What are you asked to do with the equations?

November 9th, 2016, 02:05 PM  #3 
Newbie Joined: Nov 2016 From: Estonia Posts: 2 Thanks: 0 
Solve equations and check the outcome

November 9th, 2016, 02:33 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,493 Thanks: 561 Math Focus: Wibbly wobbly timeywimey stuff. 
Okay, so let's take the first one. $\displaystyle 3~sin(x) = 2.1$ $\displaystyle sin(x) = \frac{2.1}{3}$ How do you find x from here? Dan 
November 24th, 2016, 08:52 AM  #5 
Newbie Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All 
a) 3sin x = 2.1 sin x = 2.1/3 sin x = 0.7 (use tables for finding the value) b) 2 + sin x = 0 sin x = 2 No solution. Because sin x range is [1, 1]. c) cos³x = 0.512 cos x = 0.8 (use tables for finding the value) d) cos²x  3 sin²x = 0 1  4 sin²x = 0 sin x = 1/root 2 OR sin x = 1/root 2 x = n*PI + (1)power n * (PI/4) for all n belongs to Z OR x = n*PI + (1)power n * (PI/4) for all n belongs to Z e) cos² x + 3cos x = 0 cos x(cos x + 3) = 0 cos x = 0 or cos x = 3 cos x = 0 => x = 2n*PI for all n belongs to Z cos x = 3 No solution. Because cos x range is [1, 1]. 
November 24th, 2016, 12:16 PM  #6 
Newbie Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All 
cos x = 0 => x = 2n*PI + PI/2 for all n belongs to Z cos x = 3 No solution. Because cos x range is [1, 1]. 