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

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We need to find x if cos t = 3/x and sint = 4/x

Use the relation (cos t)^2 + (sin t)^2 = 1

=> (3/x)^2 + (4/x)^2 = 1

=> 9/x^2 + 16/x^2 = 1

=> 25/x^2 = 1

=> x^2 = 25

=> x = 5 and x = -5

Both values give 3/x and 4/x lying between -1 and 1. Hence they are both admissible.

**The values of x are (-5 , 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.**