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

Explanation:

Step1: Recall the area - formula using definite integral

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), and the total area (A_{total}) from (x=-\frac{\pi}{2}) to (x = \frac{\pi}{2}) is (\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx). Since (\cos x) is an even - function ((\cos(-x)=\cos x)), (A_{total}=2\int_{0}^{\frac{\pi}{2}}\cos xdx). Using the integral formula (\int\cos xdx=\sin x + C), we have (A_{total}=2[\sin x]_{0}^{\frac{\pi}{2}}=2(1 - 0)=2).

Step2: Set up the relationship between the two sub - areas

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=A_{total}=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 for (A_1)

We know that (A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx=[\sin x]_{-\frac{\pi}{2}}^{k}=\sin k-\sin(-\frac{\pi}{2})=\sin k + 1). Since (A_1=\frac{3}{2}), we have the equation (\sin k+1=\frac{3}{2}).

Step4: Solve the equation for (k)

Subtract 1 from both sides of the equation (\sin k+1=\frac{3}{2}), we get (\sin k=\frac{3}{2}-1=\frac{1}{2}). In the interval (-\frac{\pi}{2}\leq k\leq\frac{\pi}{2}), the solution of the equation (\sin k=\frac{1}{2}) is (k=\frac{\pi}{6}).

Answer:

(\frac{\pi}{6})