What are the exact roots for #2costheta-sqrt3=0# in #0<=theta<2pi#?

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Answer 1 · 2018-07-27T06:00:56

The solutions are #S={pi/6, 11/6pi}#

The equation is

#2costheta-sqrt3=0#

#=>#, #costheta=sqrt3/2#

The solutions are

#=>#, #{(theta=pi/6),(theta=11/6pi):}#

Answer 2 · 2018-07-27T06:08:59

#color(brown)(theta = (pi/6, 11pi/6), 0 <= theta <= 2pi#

#2 cos theta - sqrt3 = 0#

#2 cos theta = sqrt3#

#cos theta = sqrt3 / 2#

#color(indigo)(theta = cos^-1 (sqrt3 / 2) = 30^@, 330^@, color(crimson)(" as cosine is positive in I and IV Quadrants."#