Answer :
Given that [tex]\(\cos(x) = \frac{3}{4}\)[/tex] and [tex]\(\tan(x) < 0\)[/tex], we need to find [tex]\(\cos(2x)\)[/tex].
First, remember the double-angle formula for cosine:
[tex]\[ \cos(2x) = 2 \cos^2(x) - 1 \][/tex]
Since [tex]\(\cos(x) = \frac{3}{4}\)[/tex], we can substitute this value into the formula.
Calculate [tex]\(\cos^2(x)\)[/tex]:
[tex]\[ \cos^2(x) = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \][/tex]
Now, substitute [tex]\(\cos^2(x)\)[/tex] into the double-angle formula:
[tex]\[ \cos(2x) = 2 \left(\frac{9}{16}\right) - 1 \][/tex]
Simplify the expression:
[tex]\[ \cos(2x) = \frac{18}{16} - 1 = \frac{18}{16} - \frac{16}{16} = \frac{2}{16} = \frac{1}{8} \][/tex]
Therefore, the value of [tex]\(\cos(2x)\)[/tex] is [tex]\(\frac{1}{8}\)[/tex].
So the correct answer is:
[tex]\[ \boxed{\frac{1}{8}} \][/tex]
First, remember the double-angle formula for cosine:
[tex]\[ \cos(2x) = 2 \cos^2(x) - 1 \][/tex]
Since [tex]\(\cos(x) = \frac{3}{4}\)[/tex], we can substitute this value into the formula.
Calculate [tex]\(\cos^2(x)\)[/tex]:
[tex]\[ \cos^2(x) = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \][/tex]
Now, substitute [tex]\(\cos^2(x)\)[/tex] into the double-angle formula:
[tex]\[ \cos(2x) = 2 \left(\frac{9}{16}\right) - 1 \][/tex]
Simplify the expression:
[tex]\[ \cos(2x) = \frac{18}{16} - 1 = \frac{18}{16} - \frac{16}{16} = \frac{2}{16} = \frac{1}{8} \][/tex]
Therefore, the value of [tex]\(\cos(2x)\)[/tex] is [tex]\(\frac{1}{8}\)[/tex].
So the correct answer is:
[tex]\[ \boxed{\frac{1}{8}} \][/tex]
