To determine whether [tex]\( \frac{5\pi}{2} \)[/tex] is a solution for the equation [tex]\( 2 \sin^2 x - \sin x - 1 = 0 \)[/tex], let's go through the relevant calculations step-by-step:
1. Substitute [tex]\( x = \frac{5\pi}{2} \)[/tex] into the equation:
We need to evaluate [tex]\( \sin \left( \frac{5\pi}{2} \right) \)[/tex].
2. Find [tex]\( \sin \left( \frac{5\pi}{2} \right) \)[/tex]:
Recall that the sine function is periodic with period [tex]\( 2\pi \)[/tex]. Specifically, [tex]\( \sin( \theta + 2\pi k) = \sin( \theta ) \)[/tex] for any integer [tex]\( k \)[/tex]. Therefore:
[tex]\[
\sin \left( \frac{5\pi}{2} \right) = \sin \left( \frac{5\pi}{2} - 2\pi \right) = \sin \left( \frac{5\pi}{2} - \frac{4\pi}{2} \right) = \sin \left( \frac{\pi}{2} \right)
\][/tex]
Since [tex]\( \sin \left( \frac{\pi}{2} \right) = 1 \)[/tex]:
[tex]\[
\sin \left( \frac{5\pi}{2} \right) = 1
\][/tex]
3. Substitute [tex]\( \sin \left( \frac{5\pi}{2} \right) = 1 \)[/tex] into the equation [tex]\( 2 \sin^2 x - \sin x - 1 = 0 \)[/tex]:
[tex]\[
2 \sin^2 \left( \frac{5\pi}{2} \right) - \sin \left( \frac{5\pi}{2} \right) - 1 = 2(1)^2 - 1 - 1 = 2 - 1 - 1 = 0
\][/tex]
4. Check the result:
The left-hand side of the equation simplifies to 0, which matches the right-hand side.
Therefore, since substituting [tex]\( x = \frac{5\pi}{2} \)[/tex] into the equation [tex]\( 2 \sin^2 x - \sin x - 1 = 0 \)[/tex] yields a true statement, the value [tex]\( \frac{5\pi}{2} \)[/tex] is indeed a solution for the equation.
Thus, the correct answer is:
A. True
