According to the information given, the trigonometric functions are given by:
[tex]\sin{2x} = \frac{12\sqrt{10}}{49}[/tex]
[tex]\cos{2x} = \frac{31}{49}[/tex]
[tex]\tan{2x} = \frac{12\sqrt{10}}{31}[/tex]
The measures of the sine and of the cosine of an angle x are related by the following equation:
[tex]\sin^2{x} + \cos^2{x} = 1[/tex]
Since [tex]\sin{x} = \frac{3}{7}[/tex]
[tex]\left(\frac{3}{7}\right)^2 + \cos^2{x} = 1[/tex]
[tex]\frac{9}{49} + \cos^2{x} = 1[/tex]
[tex]\cos^2{x} = \frac{49}{49} - \frac{9}{49}[/tex]
[tex]\cos^2{x} = \frac{40}{49}[/tex]
[tex]\cos{x} = \sqrt{\frac{40}{49}}[/tex]
[tex]\cos{x} = \frac{\sqrt{40}}{7}[/tex]
[tex]\cos{x} = \frac{2\sqrt{10}}{7}[/tex]
Now, the identities are applied:
[tex]\sin{2x} = 2\sin{x}\cos{x} = 2 \times \frac{3}{7} \times \frac{2\sqrt{10}}{7} = \frac{12\sqrt{10}}{49}[/tex]
[tex]\cos{2x} = \cos^2{x} - \sin^2{x} = \left(\frac{2\sqrt{10}}{7}\right)^2 - \left(\frac{3}{7}\right)^2 = \frac{40}{49} - \frac{9}{49} = \frac{31}{49}[/tex]
Tangent is sine divided by cosine, hence:
[tex]\tan{2x} = \frac{\sin{2x}}{\cos{2x}} = \frac{\frac{12\sqrt{10}}{49}}{\frac{31}{49}} = \frac{12\sqrt{10}}{31}[/tex]
A similar problem is given at https://brainly.com/question/13754923
