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

Explanation:

Step1: Recall the area - formula using definite integral

The area 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 from $x=-\frac{\pi}{2}$ to $x=\frac{\pi}{2}$ is $A_{total}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx$. 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 equations based on the given ratio

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 to find $k$

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

Step4: Solve the trigonometric equation for $k$

Subtract 1 from both sides of the equation $\sin k+1=\frac{3}{2}$, we get $\sin k=\frac{1}{2}$. Given that $-\frac{\pi}{2}\leq k\leq\frac{\pi}{2}$, then $k=\frac{\pi}{6}$.

Answer:

$\frac{\pi}{6}$