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 April 16th, 2018, 10:19 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 10:30 AM.
 April 16th, 2018, 10:47 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,636 Thanks: 2080 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, 10:56 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, 10:56 AM #4 Newbie   Joined: Apr 2018 From: East London Posts: 12 Thanks: 0 Thank you for the above Skip
 April 16th, 2018, 11:46 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,636 Thanks: 2080 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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