Looks more like Trig than Algebra II
They want us to apply the sum angle formula to [tex]\sin(x + \frac{5 \pi}{6})[/tex]
5π/6 is 150°; I find it easier to think in degrees.
When the tangent of x is 3/4 that's a right triangle with opposite 3, adjacent 4 so hypotenuse 5. So a sine of 3/5 and a cosine of 4/5, both positive because we're told x is in the first quadrant.
The sine sum angle formula is
[tex]\sin(a+b) = \sin a \cos b + \cos a \sin b[/tex]
[tex]\sin(x+\frac{5\pi}{6}) = \sin x \cos \frac{5\pi}{6} + \cos x \sin \frac{5\pi}{6}[/tex]
We know sin 150° = sin 30° = 1/2 and cos 150° = - cos 30° = -√3/2
[tex]\sin(x+\frac{5\pi}{6}) = \frac 3 5 (- \sqrt{3}/2) + \frac 4 5 (1/2)[/tex]
[tex]\sin(x+\frac{5\pi}{6}) = \dfrac{ 4- 3\sqrt{3}}{10}[/tex]
Third choice
