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

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cost=3/x and sint = 4/x.

To find x:

Therefore tant = sint/cost = (4/x)/(3/x) = 4/3.

Therefore tant = 4/3.

Sint = t/sqrt(1+tan^2) = (4/3)/sqrt(1+(4/3)^2) = (4/3)*3/sqrt25 = 4/5 Or -4/5.

cost = 1/sqrt(1+tan^2t) = 1/sqrt(1+(4/3)^2) = 3/5 or -3/5.

Therefore sint = 4/x = 4/5 or -4/5. So x= 5

cost = 3/x = 3/5 or -3/5. S = x = 5.

Therefore x= 5.

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.**