Trigonometry Trigonometry Math Forum

 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: 17,477 Thanks: 1313 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: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timey-wimey 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].

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