Posted by suz on .
given 7sin^2 theta + cos theta sin theta = 6 show that tan^2 theta + tan theta  6 = 0

calculus 
Reiny,
I worked it backwards, since the second equation was easier to solve
tan^2Ø + tanØ6=0
(tanØ2)(tanØ+3) = 0
tanØ = 2 or tanØ=3
case1: tanØ=2, Ø could be in I or III
In I, sinØ = 2/√5 and cosØ = 1/√5
test in first equation
LS = 7(4/5) + (1/√5)(2/√5)
= 28/5 + 2/5 = 30/5 = 6 = RS
In III
sinØ = 2/√5 , cosØ = 1/√5
LS = 7(4/5) + (2/√5)(1/√5) = 30/5 = 6 = RS
Case 2: tanØ = 3, Ø could be in II or IV
in II , sinØ = 3/√10 , cosØ = 1/√10
testing in first equation,
LS = 7(9/10) + (1/√10)((3/√10)
= 63/10  3/10 = 60/10 = 6 = RS
in IV, sinØ = 3/√10 , cosØ = 1/√10
LS = 7(9/10) + (1/√10)(3/√10)
= 63/10  3/10 = 60/10 = 6 = RS
so all 4 values of Ø satisfy both equations.