Stuck on this, don't know where to start

$\displaystyle 2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ} $

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- Feb 25th 2009, 06:33 AMBeardDouble Angle Identity
Stuck on this, don't know where to start

$\displaystyle 2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ} $ - Feb 25th 2009, 06:41 AMHallsofIvy
- Feb 25th 2009, 06:43 AMChop Suey
- Feb 25th 2009, 06:44 AMADARSH
- Feb 25th 2009, 06:45 AMBeard
at 45

- Feb 25th 2009, 06:47 AMChop Suey
I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.

- Feb 25th 2009, 06:53 AMADARSH
Read the initial lines

This is an equation which has solutions (Nod) and not identity(Shake) - Feb 25th 2009, 06:53 AMHallsofIvy
- Feb 25th 2009, 07:44 AMBeard