the region bounded by the x - axis and the part of the graph of (y = cos x) between (x=-\frac{pi}{2}) and (x…

Explanation:

Step1: Recall the area - under - curve formula

The area (A) under the curve (y = f(x)) from (x=a) to (x = b) is given by (A=\int_{a}^{b}|f(x)|dx). Here, (y = \cos x) is non - negative on the interval (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), so the area under the curve (y=\cos x) from (x =-\frac{\pi}{2}) to (x=\frac{\pi}{2}) is (A=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx). Using the property (\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx) for an even function (f(x)) (and (\cos x) is an even function, i.e., (\cos(-x)=\cos x)), we have (\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx = 2\int_{0}^{\frac{\pi}{2}}\cos xdx=2[\sin x]_0^{\frac{\pi}{2}}=2).

Step2: Set up the equation based on the area relationship

Let (A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx) and (A_2=\int_{k}^{\frac{\pi}{2}}\cos xdx). We know that (A_1 = 3A_2), and also (A_1+A_2=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx = 2). Substituting (A_1 = 3A_2) into (A_1+A_2 = 2), we get (3A_2+A_2=2), so (4A_2 = 2), then (A_2=\frac{1}{2}) and (A_1=\frac{3}{2}).

Step3: Calculate the integral (\int_{-\frac{\pi}{2}}^{k}\cos xdx)

We know that (\int_{-\frac{\pi}{2}}^{k}\cos xdx=\left[\sin x\right]{-\frac{\pi}{2}}^{k}=\sin k-\sin\left(-\frac{\pi}{2}\right)=\sin k + 1). Since (\int{-\frac{\pi}{2}}^{k}\cos xdx=\frac{3}{2}), we have the equation (\sin k+1=\frac{3}{2}).

Step4: Solve for (k)

Subtract 1 from both sides of the equation (\sin k + 1=\frac{3}{2}), we get (\sin k=\frac{1}{2}). Since (k\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), then (k=\frac{\pi}{6}).

Answer:

C. (\frac{\pi}{6})