g(v)*2cos theta+1=0

The solution set of (5+4cos theta)(2cos theta+1)=0 in the interval [0,2 pi] is

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

If (cos theta_(1))/(cos theta_(2))+(sin theta_(1))/(sin theta_(2))=(cos theta_(0))/(cos theta_(2))+(sin theta_(0))/(sin theta_(2))=1 , where theta_(1) and theta_(0) do not differ by can even multiple of pi , prove that (cos theta_(1)*cos theta_(0))/(cos^( 2)theta_(2))+(sin theta_(1)*sin theta_(0))/(sin^(2) theta_(2))=-1

If 2cos^(2)theta + cos theta -1 =0 then theta =

If cos 6theta + cos 4theta + cos 2theta +1 =0 for 0 le theta le pi then theta =

If 0 le theta le pr, cos 6theta + cos 4theta + cos 2theta + 1 =0 then the ascending order of the values of theta is

If : cos 6 theta + cos 4 theta + cos 2 theta + 1 = 0 , where, 0 lt theta lt 180^(@) , then : theta =

The value of cos^(2)((pi)/(6)+theta)-sin^(2)((pi)/(6)-theta) is (1)/(2)cos2 theta b.0c*-(1)/(2)cos2 theta d.-cot35^(0)

The solution of sin 2 theta + cos 2theta + sin theta + cos theta + 1 =0 in the first quadrant is