5. let $f(x)=5sin(2x)-5cos x$. find all solutions to the equation $f(x)=0$ on the interval $0,pi$.

Explanation:

Step1: Use double - angle formula

Recall $\sin(2x)=2\sin x\cos x$. So $f(x)=5\times2\sin x\cos x - 5\cos x=0$.

Step2: Factor out common factor

Factor out $5\cos x$ from the left - hand side of the equation: $5\cos x(2\sin x - 1)=0$.

Step3: Set each factor equal to zero

Set $5\cos x = 0$ and $2\sin x - 1=0$. For $5\cos x = 0$, we have $\cos x = 0$. In the interval $[0,\pi]$, $x=\frac{\pi}{2}$. For $2\sin x - 1 = 0$, we get $\sin x=\frac{1}{2}$. In the interval $[0,\pi]$, $x = \frac{\pi}{6}$ or $x=\frac{5\pi}{6}$.

Answer:

$x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$