April 16th, 2018, 11:19 AM  #1 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0  Trig double angles
Sin (x+60) = 2 Sinx Solve for x Last edited by Vee88; April 16th, 2018 at 11:30 AM. 
April 16th, 2018, 11:47 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
sin(x)cos(60°) + cos(x)sin(60°) = 2sin(x) tan(x)/2 + √3/2 = 2tan(x) sqrt(3) = 3tan(x) tan(x) = 1/√3 x = (30 + 180k)°, where k is an integer. 
April 16th, 2018, 11:56 AM  #3 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0  General Solution Trig
Determine the General Solution 4Cos^2 (x)  Sin (2x 180)= Tan^2 675 
April 16th, 2018, 11:56 AM  #4 
Newbie Joined: Apr 2018 From: East London Posts: 12 Thanks: 0 
Thank you for the above Skip

April 16th, 2018, 12:46 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
4cos²(x)  sin(2x  180°)= tan²(675$^\circ$) 4cos²(x) + sin(2x) = 1 4cos²(x) + 2sin(x)cos(x) = cos²(x) + sin²(x) sin²(x)  2sin(x)cos(x) + cos²(x) = 4cos²(x) (sin(x)  cos(x))² = (2cos(x))² sin(x)  cos(x) = ±2cos(x) sin(x) = cos(x) ± 2cos(x) = cos(x) or 3cos(x) tan(x) = 1 or 3 x = (135 + 180k)$^\circ$ or arctan(3) + 180k$^\circ$ where (by use of a calculator) arctan(3) = 71.565051177...$^\circ$ and k is an integer. 

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angles, double, trig 
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