User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 November 19th, 2015, 12:47 PM #1 Senior Member   Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 General Solution I seem to be getting this wrong. Obtain the general solution of the equation $\displaystyle Sin(2x)=Sin(x)$ $\displaystyle 2CosxSinx-Sinx=0$ $\displaystyle Sinx(2cosx-1)=0$ $\displaystyle Sinx=0$ PV = 0 General Solution: $\displaystyle x=pi*k + 0$ $\displaystyle Cosx=\frac{1}{2}$ PV = 60 Degrees General Solution: $\displaystyle x = 360*k +- 60$ k being any integer November 19th, 2015, 01:16 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,837 Thanks: 1479 in the interval $0 \le x < 360^\circ$ , $\cos{x} = \dfrac{1}{2}$ at $x = 60^\circ$ and $x = 300^\circ$ general ... x = 60 + 360k x = 300 + 360k why the switch from radians to degrees for the two zero equations? Thanks from topsquark Tags general, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post woo Differential Equations 1 April 27th, 2015 04:51 PM Tooperoo Calculus 4 October 16th, 2013 01:10 AM hoohah2 Algebra 4 August 11th, 2013 12:58 PM AzraaBux Algebra 4 May 31st, 2013 07:13 AM mathbalarka Calculus 11 May 5th, 2013 11:01 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      