If we have tan x = √3/3, I would expect ...

If we have
tan x = √3/3, and we know
tan x = sin x / cos x, then I would expect
sin x = √3, and
cos x = 3.

But this is not the case, as will be proven if you solve the problem using other trigonometric identities.

I'm perplexed. Is my knowledge of algebra flawed?
Could someone please explain to me why we can't simply 'substitute', i.e. sin x = √3 ?

The first thing that should tell you there is a problem with your rationale is that sine and cosine are never greater than 1 nor less than -1. So, sinx cannot be √3 because √3 > 1. And, likewise cosx cannot be 3 because 3 > 1.

Consider a right triangle with legs a & b and hypotenuse c and let x be the angle opposite side 'a'. Then sinx=a/c and cosx=b/c so that a/c=√3 and b/c=3 for some a, b, and c ... Notice that it is not that sinx=√3 but rather it is a/c that is √3. Similarly, it is not that cosx=3 but rather it is b/c that is 3.

If you mark side 'a' of the right triangle as √3 and mark side b as 3, then you can see how tanx, √3, and 3 are related ... tanx = (√3)/3 because tanx = opposite/adjacent.

If you now apply the Pythagorean Theorem to the triangle you get that the hypotenuse is √12 or 2√3. Now, from this triangle you can determine the value of all the trig functions for angle x.

Hope that helps.
Have a good one!
.

Your trigonometry has to use more common sense. Remember that every sine and cosine value is less than 1.
sin x can never equal √3 or 1.73205
cos x can never equal 3
That is why you can't take inverse trigonometric values greater than 1 or less than -1.

To do this, use the ratio of √3 and 3 to make new values to be for sin x/cos x.
√3 / 3 = 1/√3
Multiply 1/2 on the numerator and denominator.
(1/2) / (√3/2)
Now familiar ratios are found, from corresponding angle values of sine and cosine. You can now do what you wanted.