Trig Functions — Question 4

Question 4 - Solution

Simplify. Double-angle identities give

$$\boxed{f(x)=\frac12+\frac{\sqrt2}{2}\sin(2x+\pi/4)}.$$

The sine attains both $-1$ and $1$ on the given interval, so

$$\boxed{f_{\min}=\frac{1-\sqrt2}{2},\qquad f_{\max}=\frac{1+\sqrt2}{2}}.$$

Solve without losing roots. Subtracting $1$ from the original expression gives

$$f(x)-1=\sin x\cos x-\sin^2x=\sin x(\cos x-\sin x).$$

Thus $\sin x=0$ or $\cos x=\sin x$. Restricting both families to $[0,2\pi]$ gives

$$\boxed{x=0,\ \pi/4,\ \pi,\ 5\pi/4,\ 2\pi}.$$

Each listed value makes one factor zero, so each satisfies the original equation.