Area under the trignometric curve!

So $a$ is a value where $f(a) = 0$ (also the 1st positive value that does this). We can predict some values of $a$ by writing out the function in a factored form: $f(x) = sin(2x) - \sqrt{3}sin(x) \\ f(x) = 2 \dot \ sinx \dot \ cosx - \sqrt{3}sinx \\ f(x) = sinx(2\dot \ cosx - \sqrt{3})$

Since $f(a) = 0$ , either $sina = 0$ , or $2\dot \ cosa - \sqrt{3} = 0$ . From the first condition (??), $a$ would have to be $\pi$ . From the second condition, $cosa = \frac{\sqrt{3}} {2}$ , by some manipulation. So $a$ would be $\pi / 6$ . But between these 2 options, $\pi / 6$ is smaller, so it's the appropriate value for $a$ . $$

Now let's find an expression for the area, $A$ , in terms of $a$ , which is the integral $A = \int_0^a (\! 2\dot \ sinx \dot \ cosx - \sqrt{3} \dot \ sinx) \ \mbox{d}x$ . Solving the integral gives us $1 - \sqrt{3} - cos^2a + \sqrt{3}\dot \ cosa$ (hopefully that's right :P). $$

Plugging in this expression for $A$ into $4A + 8\dot \ cosa$ , then changing $a$ to $\pi / 6$ and simplifying gives us the correct answer of $7$ .

a=pi/6 .. !! one thing to notice is that A is negative .. Everything else is simple integration .. !! the limits for integration are from 0 to pi/6