Question

How do you find the exact value of `3cos^2theta-costheta-2=0` in the interval `0<=theta<2pi`?

Answer 1

Solution : In interval `0 <= theta < 2 pi ; theta =0^0 , theta = 131.81^0 , theta = 228.19^0`

Explanation:

Note : Assumed, possible values of `theta` are asked to find out.

`3cos^2 theta -cos theta -2=0 or 3cos^2 theta -3cos theta +2cos theta-2=0 or 3cos theta(cos theta-1) +2(cos theta -1)=0 or (3cos theta+2)(cos theta-1) =0 :.`. Either #3cos theta +2=0 :. cos theta = -2/3 or cos theta-1=0 :. cos theta =1# In interval `0 <= theta < 2 pi ; cos 0 =1 :. theta = 0^0 ; cos theta = -2/3 :. theta =cos^-1 (-2/3) = 131.81^0(2dp)`Also `theta =180+(180-131.81)=228.19` since `cos 228.19 = -2/3`

Solution : In interval `0 <= theta < 2 pi ; theta =0^0 , theta = 131.81^0 , theta = 228.19^0` [Ans]