savb517 savb517

20-02-2017
Mathematics

contestada

solve the trigonometric equation
3-3cosx=2 sin^2 x

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Amena2000 Amena2000

25-02-2017

= 3 - 3cos x - 2(1- cos^2 x)
= 2cos^2 x - 3cos x + 1
=(cos x -1)(cosx - 1/2)
now cos x = 1 or cos x= 1/2
draw a graph we get = + ve values
hence .. 0 ,180,360 and 60,240, degrees in range x<2pi

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