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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Originally Posted by Beard 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} $ Sorry, I thought you were trying to prove an identity!
Originally Posted by Beard 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} $ I'll start you off: $\displaystyle 2\cos^{2}x = 2\sin(x)\cos(x) + 1 \implies 2\cos^{2}x -1 = 2\sin(x)\cos(x) \implies \cos{2x} = \sin{2x}$ When does the cosine of an angle equals the sine of an angle?
Originally Posted by Beard 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} $ First of all this is not an identity Put x= 0 In case you need to solve the equation Put $\displaystyle 1 = sin^2(x) + cos^2(x)$ EDIT: Far too late
at 45
I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.
Read the initial lines This is an equation which has solutions and not identity
Originally Posted by Beard at 45 Sine and cosine are equal for 45 degrees and at 180+ 45= 225 degrees. That, of course, would be 2x. Now what values does x have?
Originally Posted by Chop Suey I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation. Sorry about this
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