If $\sin^2 \theta + 3\cos\theta - 2 = 0$, find the value of $\cos^3 \theta + \sec^3 \theta$.
+128406 sin^2x + 3cos x - 2 = 0
( 1 - cos^2x) + 3cosx - 2 = 0
-cos^2x + 3cos x - 1 = 0
cos^2 x - 3cosx + 1 = 0 let cos x = a
a^2 - 3a = -1 completete the square on a
a^2 - 3a + 9/4 = -1 + 9/4
(a - 3/2)^2 = 5/4 take both roots
a - 3/2 = sqrt (5)/2 a - 3/2 = -sqrt (5)/2
a = sqrt (5)/2 + 3/2 a = 3/2 - sqrt (5)/2
cos x = sqrt (5)/2 + 3/2 cos x = (3 - sqrt 5) / 2
Not possible
cos^3x + sec^3x =
(cos x + sec x) ( cos^2 x - cosxsecx + sec^2x) =
(cos x + secx) ( cos^2 x + sec^2x - 1)
[ (3 - sqrt 5)/2 + 2/ (3 -sqrt 5) ] [ ( ( 3 -sqrt 5)/2 )^2 + (2/(3 -sqrt 2))^2 - 1 ] = 18
