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: Calculate the total area

The area of the region bounded by the (x -)axis and (y = \cos x) between (x=-\frac{\pi}{2}) and (x = \frac{\pi}{2}) is (A=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\vert\cos x\vert dx). Since (\cos x\geqslant0) when (x\in[-\frac{\pi}{2},\frac{\pi}{2}]), then (A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx). According to the integral formula (\int\cos xdx=\sin x + C), we have (A=\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 area - ratio equation

Let the area of the region for (-\frac{\pi}{2}\leqslant x\leqslant k) be (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), and the area of the region for (k\leqslant x\leqslant\frac{\pi}{2}) be (A_2=\int{k}^{\frac{\pi}{2}}\cos xdx=\left[\sin x\right]_{k}^{\frac{\pi}{2}}=1-\sin k). Given that (A_1 = 3A_2), so (\sin k + 1=3(1 - \sin k)).

Step3: Solve the equation for (\sin k)

Expand the right - hand side of the equation: (\sin k+1 = 3-3\sin k). Move the terms with (\sin k) to one side: (\sin k + 3\sin k=3 - 1). Combine like terms: (4\sin k=2). Solve for (\sin k): (\sin k=\frac{1}{2}).

Step4: Find the value of (k)

Since (k\in[-\frac{\pi}{2},\frac{\pi}{2}]) and (\sin k=\frac{1}{2}), then (k=\frac{\pi}{6}).

Answer:

(\frac{\pi}{6})