Question

How do you do this question?

Answer 1

Answer:

C) π/6

Step-by-step explanation:

The area under the curve from x=-π/2 to x=k is 3 times the area under the curve from x=k to x=π/2.

$\int\limits^k_{-\frac{\pi }{2}} {cos\ x} \, dx = 3\int\limits^{\frac{\pi }{2}}_k {cos\ x} \, dx\\sin\ x\ |^{k}_{-\frac{\pi }{2}} = 3\ sin\ x\ |^{\frac{\pi }{2}}_k\\(sin\ k - (-1)) = 3\ (1 - sin\ k)\\sin\ k + 1 = 3 - 3\ sin\ k\\4\ sin\ k = 2\\sin\ k = \frac{1}{2}\\k = \frac{\pi}{6}$

Graph: desmos.com/calculator/mezlen9hb4