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

Explanation:

Step1: Use the integral formula for the area

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). For (y = \cos x), (\int\cos xdx=\sin x + C). The total area from (x =-\frac{\pi}{2}) to (x=\frac{\pi}{2}) is (A_{total}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx=\left[\sin x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\sin\frac{\pi}{2}-\sin(-\frac{\pi}{2})=1 - (- 1)=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 (A_1+A_2=A_{total}). Substituting (A_1 = 3A_2) into (A_1+A_2=2), we get (3A_2+A_2=2), so (A_2=\frac{1}{2}) and (A_1=\frac{3}{2}). Also, since (A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx=\left[\sin x\right]_{-\frac{\pi}{2}}^{k}=\sin k-\sin(-\frac{\pi}{2})=\sin k + 1).

Step3: Solve for (k)

Set (\sin k+1=\frac{3}{2}). Then (\sin k=\frac{3}{2}-1=\frac{1}{2}). We know that (y = \sin x), and for (x\in[-\frac{\pi}{2},\frac{\pi}{2}]), when (\sin k=\frac{1}{2}), (k=\frac{\pi}{6}).

Answer:

(\frac{\pi}{6})