Of course, let's solve this step-by-step.
Given the equation:
[tex]\[ 5 \cos A + 3 = 0 \][/tex]
We can solve for [tex]\(\cos A\)[/tex]:
[tex]\[ 5 \cos A = -3 \][/tex]
[tex]\[ \cos A = -\frac{3}{5} \][/tex]
Since [tex]\(180^{\circ}
To find [tex]\(\sin A\)[/tex], we use the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Substitute the value of [tex]\(\cos A\)[/tex]:
[tex]\[ \sin^2 A + \left( -\frac{3}{5} \right)^2 = 1 \][/tex]
[tex]\[ \sin^2 A + \frac{9}{25} = 1 \][/tex]
[tex]\[ \sin^2 A = 1 - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 A = \frac{25}{25} - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 A = \frac{16}{25} \][/tex]
Taking the square root, we have:
[tex]\[ \sin A = \pm \frac{4}{5} \][/tex]
Since [tex]\(A\)[/tex] is between [tex]\(180^{\circ}\)[/tex] and [tex]\(360^{\circ}\)[/tex], which means it's either in the third quadrant or the fourth quadrant. In these quadrants, cosine is negative and sine is negative. Thus:
[tex]\[ \sin A = - \frac{4}{5} \][/tex]
Now, we find [tex]\(\sin A + \cos A\)[/tex]:
[tex]\[ \sin A + \cos A = -\frac{4}{5} + \left( -\frac{3}{5} \right) \][/tex]
[tex]\[ \sin A + \cos A = -\frac{4}{5} - \frac{3}{5} \][/tex]
[tex]\[ \sin A + \cos A = -\frac{7}{5} \][/tex]
So the correct value of [tex]\(\sin A + \cos A\)[/tex] is [tex]\(-\frac{7}{5}\)[/tex].
Thus, the answer is:
c. [tex]\(\boxed{-\frac{7}{5}}\)[/tex]
