# If it is ggiven that sin2x + 3cosx-2 = 0, then find the value of cos3x + sec3x

Arun
25757 Points
5 years ago
Dear Student
sin^2x = 1 – cos^2 x
now
1 – cos^2 x + 3 cos x – 2=0
cos^2 x- 3cos x +1 = 0
let cos x = t
then it will form a quadratic equation
t^2 – 3t + 1 = 0
t = (3 + $\sqrt$5)/2  and t = (3 –  $\sqrt$5)/2
1st value is cancelled
now
cos x= (3 –  $\sqrt$5)/2
now you can easily find out the answer.

regards
Rishabh Sankla
33 Points
3 years ago
Dear Student,

$sin^2 x + 3 cos x = 2$

$\Rightarrow (1-cos^2 x) + 3 cos x = 2$

$\Rightarrow cos^2 x - 3 cos x = -1$                [After shifting terms]

$\Rightarrow cos x(cos x - 3) = -1$

$\Rightarrow cos x - 3 = -sec x$

$\Rightarrow cos x + sec x = 3$

$\Rightarrow (cos x + sec x)^3 = (3)^3$

$\Rightarrow cos^3 x + sec^3 x + 3 cos x sec x(cos x + sec x) = 27$

$\Rightarrow cos^3 x + sec^3 x + 3\times3 = 27$

$\Rightarrow cos^3 x + sec^3 x = 18$

This is the simpler method to find answer to this question.

Regards
Rishabh