Suppose that x > 0 , cos(θ) = 45/x and sin(θ) = 24/x . Find x

First, we'll impose the constraint of existence of the
trigonometric functions sine and cosine:

-1=< sin(θ)
=<1

sin(θ) = 24/x 

-1=<
24/x =<1

We'll multiply both sides by
x:

-x=< 24 =<
x

-1=< cos(θ) =<1

cos(θ)
= 45/x

-1=< cos(θ)
=<1

-1=< 45/x
=<1

We'll multiply both sides by
x:

-x=< 45 =< x

From both
inequalities, we'll get the interval for adissible value for x: [45 ;
+infinite)

Now, we'll solve the equtaion, applying the fundamental
formula of trigonomtery:

[sin(θ)]^2 + [cos(θ)]^2 
=1

(576+2025)/x^2  = 1

We'll multiply
both sides by x^2:

2601 = x^2

We'll
apply square root both sides:

x =
51

x =
-51

Since just 51 is in the interval of
admissible values, we'll accept just a single solution x =
51.