The region bounded by the x-axis and the part of the graph of y = cos x between x = −π/2 and x = π/2 is separated into two regions by the line x = k. If the area of the region for -π/2 ≤ x ≤ k is three times the area of the region for k ≤ x , then k =

The problem is asking for the value of k that divides the area under the curve y = cos x from x = -π/2 to x = π/2 into two parts such that one part is three times the other.

The area under the curve y = cos x from x = a to x = b is given by the integral ∫ from a to b of cos x dx, which equals [sin x] from a to b, or sin b - sin a.

We know that the total area under the curve from x = -π/2 to x = π/2 is sin(π/2) - sin(-π/2) = 1 - (-1) = 2.

Let A1 be the area of the region for -π/2 ≤ x ≤ k and A2 be the area of the region for k ≤ x ≤ π/2. We are given that A1 = 3A2.

Since A1 + A2 = 2, we can substitute A1 = 3A2 into this equation to get 3A2 + A2 = 2, or 4A2 = 2. Solving for A2 gives A2 = 2/4 = 0.5.

Then A1 = 3A2 = 3*0.5 = 1.5.

So we need to find k such that the area under the curve from x = -π/2 to x = k is 1.5. This means we need to solve the equation ∫ from -π/2 to k of cos x dx = 1.5 for k.

The integral ∫ from -π/2 to k of cos x dx equals sin k - sin(-π/2) = sin k + 1.

Setting this equal to 1.5 and solving for sin k gives sin k = 1.5 - 1 = 0.5.

The value of k that satisfies this equation in the interval -π/2 ≤ k ≤ π/2 is k = π/6.