We have `sqrt(3) sin x = 2 (cos x + cos^(2) x)` ...(i) Squaring both sides, we get `3 sin^(2) x=4(cos^(2)x+cos^(4)x+2 cos^(3) x)` or `3(1-cos^(2)x)=4 cos^(2)x+4 cos^(4) x+8 cos^(3) x`. or `4 cos^(4) x+8 cos^(3) x+7 cos^(2) x-3 =0` or `(cos x+1)(2 cos x-1) (2 cos^(2) x+3 cos x+3)=0` For `cos x=-1, x=(2n+1) pi, n in Z` Clearly, `cos x=-1` satisfies eq. (i). Now, `cos x=1/2` Clearly, for `cosx=1/2` from eq. (i), `sin x=sqrt(3)/2` So, x lies in `1^(st)` quadrant. `:. x=2npi +pi/3, n in Z`
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Let alpha be a root of the equation (2sinx-cosx)(1+cosx)=sin^2x beta is a root of the equation 3cos^2x-10cosx+3=0 gamma be a root of the equation 1-sin2x=cosx-sinx 0<=alpha,beta,gamma<=pi/2 cos alpha+cos beta+cos gamma can be equal to (A) 3sqrt(6)+2sqrt(2)+6sqrt(2) (B) (3sqrt(3)+8)/12 (C) (3sqrt(3)+2)/6 (D) none of these
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