Mathematics - Trigonometric Equation Question with Solution | TestHub

The inequation is $\sin x+\sin 2x+\sin 3x>0$. Using sum-to-product formula for $\sin x+\sin 3x=2\sin 2x\cos x$, we get $2\sin 2x\cos x+\sin 2x>0$. Factoring, $\sin 2x(1+2\cos x)>0$. This requires two cases: Case 1:$\sin 2x>0$ AND $1+2\cos x>0⟹\cos x>-1/2$. In$[0,2\pi )$: $\sin 2x>0⟹x\in (0,\pi /2)\cup (\pi ,3\pi /2)$. $\cos x>-1/2⟹x\in [0,2\pi /3)\cup (4\pi /3,2\pi )$. The intersection is $(0,\pi /2)\cup (4\pi /3,3\pi /2)$. Case 2:$\sin 2x<0$ AND $1+2\cos x<0⟹\cos x<-1/2$. In$[0,2\pi )$: $\sin 2x<0⟹x\in (\pi /2,\pi )\cup (3\pi /2,2\pi )$. $\cos x<-1/2⟹x\in (2\pi /3,4\pi /3)$. The intersection is $(2\pi /3,\pi )$.

P: The solution set in $[0,2\pi )$ is $(0,\pi /2)\cup (2\pi /3,\pi )\cup (4\pi /3,3\pi /2)$. The total length is $(\pi /2-0)+(\pi -2\pi /3)+(3\pi /2-4\pi /3)=\pi /2+\pi /3+\pi /6=\pi$. (P)-(4). Q: The solution set consists of 3 disjoint intervals. (Q)-(3). R: For$\sin 2x(1+2\cos x)=0$, we have $\sin 2x=0$ or $\cos x=-1/2$. Solutions are $x=n\pi /2$ or $x=2n\pi \pm 2\pi /3$. The smallest positive solution is $\pi /2$. (R)-(1). S: The solution set in$[0,\pi )$ is $(0,\pi /2)\cup (2\pi /3,\pi )$. The total length is $(\pi /2-0)+(\pi -2\pi /3)=\pi /2+\pi /3=5\pi /6$. (S)-(2).