first derivatives test and trig functions

I am supposed to use the first derivative test to find any local min/max of $\displaystyle f(x)=sin^2(x)+cos)(x)$ on the interval $\displaystyle \frac{\pi}{6},\frac{3\pi}{2}$

Take the derivative and get $\displaystyle f'(x)=2cos(x)-sin(x)$

Set the derivative = 0 and I get $\displaystyle 2cos(x)=sin(x)$

I graphed both 2cos(x) and sin(x) and there are 2 intersections. How do I solve for x?

Thanks!

Re: first derivatives test and trig functions

Quote:

Originally Posted by

**baldysm** I am supposed to use the first derivative test to find any local min/max of $\displaystyle f(x)=sin^2(x)+cos)(x)$ on the interval $\displaystyle \frac{\pi}{6},\frac{3\pi}{2}$

Take the derivative and get $\displaystyle \color{red}f'(x)=2cos(x)-sin(x)$

You have done the derivative incorrectly.

$\displaystyle f'(x)=2\sin(x)\cos(x)-\sin(x)$

Re: first derivatives test and trig functions

Hey baldysm.

Hint: sin(x)/cos(x) = tan(x).

Edit: Also check HallsOfIvy's post above.