What is x if cos t = 3/x and sint = 4/x ?

We'll impose the conditions of existence of the
trigonometric functions sine and cosine:

-1=< sin(t)
=<1

sin(t) =
4/x 

-1=<
4/x =<1

We'll multiply both sides by
x:

-x=< 4 =<
x

-1=< cos(t)
=<1

cos(t) =
3/x

-1=< cos(t)
=<1

-1=< 3/x
=<1

We'll multiply both sides by
x:

-x=< 3 =<
x

From both inequalities, we'll get the interval for
adissible value for x: [4 ; +infinite)

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

[sin(t)]^2 + [cos(t)]^2 
=1

(9+16)/x^2  = 1

We'll
multiply both sides by x^2:

25 =
x^2

We'll apply square root both
sides:

x =
5

x =
-5

Since just 5 is in the
interval of admissible values, we'll reject the negative value x = -5. So, the only
solution is x = 5.