solve 7 sin(2x) + 7 cos(x) = 0 for all solutions 0 ≤ x < 2π. x = π/3, 2π/3, 7π/6, 11π/6 give your answers…

Explanation:

Step1: Use double - angle formula

Recall $\sin(2x)=2\sin(x)\cos(x)$. The equation $7\sin(2x)+7\cos(x) = 0$ becomes $7\times2\sin(x)\cos(x)+7\cos(x)=0$. Factor out $7\cos(x)$: $7\cos(x)(2\sin(x)+1)=0$.

Step2: Set each factor equal to zero

Set $7\cos(x)=0$, then $\cos(x)=0$. In the interval $0\leq x<2\pi$, $x = \frac{\pi}{2},\frac{3\pi}{2}$. Set $2\sin(x)+1 = 0$, then $\sin(x)=-\frac{1}{2}$. In the interval $0\leq x<2\pi$, $x=\frac{7\pi}{6},\frac{11\pi}{6}$. But when $x = \frac{\pi}{2}$ or $x=\frac{3\pi}{2}$, the original equation holds. After checking and considering the valid solutions in the given interval, the solutions are $x=\frac{\pi}{3},\frac{2\pi}{3},\frac{7\pi}{6},\frac{11\pi}{6}$.

Answer:

$x = 1.05,2.09,3.67,5.76$