Consider the cubic equation `x^3-(1+cos theta+sin theta)x^2+(cos theta sin theta+cos theta+sin theta)x-sin theta. cos theta =0` Whose roots are `x_1, x_2 and x_3`

Consider the cubic equation x^(3)-(1+cos theta+sin theta)x^(2)+(cos theta sin theta +cos theta + sin theta)x - sin theta cos theta =0 . When roots are x_(1),x_(2) and x_(3) . The value of x_(1)^(2)+x_(2)^(2)+x_(3)^(2) equals

cosider the cubic equation : x^(3)-(1+cos theta+sin theta)x^(2)+(cos theta sin theta+cos theta+sin theta)x-sin theta cos theta=0 whose roots are x_(1),x_(2),x_(3) .The value of (x_(1))^(2)+(x_(2))^(2)+(x_(3))^(2) equals

Consider the cubic equation x^(3)-(1+cos theta+sin theta)x^(2)+(cos theta sin theta +cos theta + sin theta)x - sin theta cos theta =0 . When roots are x_(1),x_(2) and x_(3) . Number of values of theta in [0,2pi] for which at least two roots are equal

8sin theta cos theta cos2 theta cos4 theta=sin x the x=

(sin theta+cos theta)(1-sin theta cos theta)=sin^3 theta+cos^3 theta

Prove that (cos^3 theta-sin^3 theta)/(cos theta-sin theta) = 1+cos theta sin theta

8sin theta cos theta cos2 theta cos4 theta=sin x rArr x=

Prove that: (sin^(3) theta + cos^(3) theta)/(sin theta + cos theta) = 1-sin theta cos theta

8sin theta cos theta*cos2 theta cos4 theta=sin x rArr x=

3. sin^(3)theta+cos^(3)theta=(sin theta+cos theta)(1-sin theta cos theta)