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The trick here is to recognize that "sin omega = 3/4" provides us with the length of the opposite side of a first-quadrant triangle and the length of the hypotenuse; they are 3 and 4 respectively.

Find the length of the adjacent side (x) using the Pythagorean Theorem:  3^2 + x^2 = 4^2, or x^2 = 16 - 9 = 7.  Then the length of the adj. side, x, is sqrt(7).

The cosine of omega is  adj / hyp, or sqrt(7) / 4.

Answer: The answer is [tex]\cos \theta=\dfrac{\sqrt{7}}{4}.[/tex]

Step-by-step explanation:  Given that

[tex]\sin \theta=\dfrac{3}{4}[/tex] and [tex]\theta[/tex] is in Quadrant I.

We are to find the value of [tex]\cos \theta.[/tex]

We have the following trigonometric identity :

[tex]\sin^2\theta+\cos^2\theta=1\\\\\Rightarrow \cos^2\theta=1-\sin^2\theta\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\sin^2\theta}\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\left(\dfrac{3}{4}\right)^2}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\dfrac{9}{16}}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{\dfrac{7}{16}}\\\\\\\Rightarrow \cos \theta=\pm\dfrac{\sqrt{7}}{4}.[/tex]

Since [tex]\theta[/tex] lies in the first quadrant, so cosine of [tex]\theta[/tex] will be positive.

Thus,

[tex]\cos \theta=\dfrac{\sqrt{7}}{4}.[/tex]

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