Mathematics - Trigonometric Equation Question with Solution | TestHub

The given inequality is $\sin x+\sin 2x>0$. We can rewrite $\sin 2x$ as $2\sin x\cos x$. So, $\sin x+2\sin x\cos x>0⟹\sin x(1+2\cos x)>0$. We need to analyze two cases for the product to be positive: Case 1: $\sin x>0$ and $1+2\cos x>0$.$\sin x>0⟹x\in (0,\pi )$ in the interval $[0,2\pi ]$.$1+2\cos x>0⟹\cos x>-1/2$. In $[0,2\pi ]$, this means $x\in [0,2\pi /3)\cup (4\pi /3,2\pi ]$. The intersection of these conditions is $x\in (0,2\pi /3)$. Case 2: $\sin x<0$ and $1+2\cos x<0$.$\sin x<0⟹x\in (\pi ,2\pi )$ in the interval $[0,2\pi ]$.$1+2\cos x<0⟹\cos x<-1/2$. In $[0,2\pi ]$, this means $x\in (2\pi /3,4\pi /3)$. The intersection of these conditions is $x\in (\pi ,4\pi /3)$.

Combining the solutions from Case 1 and Case 2, the set of all values of $x\in [0,2\pi ]$ for which $\sin x+\sin 2x>0$ is $(0,2\pi /3)\cup (\pi ,4\pi /3)$.