solve the equation on the interval $0,2pi)$ \n $sin 2x = -sqrt{3}sin x$ \n select the correct choice below…

solve the equation on the interval $0,2pi)$ \n $sin 2x = -sqrt{3}sin x$ \n select the correct choice below and, if necessary, fill in the answer box to complete your choice \n a. the solution set is \n (type your answer in radians. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.) \n b. the solution is the empty set

solve the equation on the interval $0,2pi)$ \n $sin 2x = -sqrt{3}sin x$ \n select the correct choice below and, if necessary, fill in the answer box to complete your choice \n a. the solution set is \n (type your answer in radians. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.) \n b. the solution is the empty set

Answer

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula (\sin2x = 2\sin x\cos x). The given equation (\sin2x=-\sqrt{3}\sin x) becomes (2\sin x\cos x=-\sqrt{3}\sin x).

Step2: Move all terms to one side

Subtract (-\sqrt{3}\sin x) from both sides: (2\sin x\cos x+\sqrt{3}\sin x = 0). Factor out (\sin x): (\sin x(2\cos x+\sqrt{3}) = 0).

Step3: Set each factor equal to zero

  • Case 1: (\sin x = 0) On the interval ([0,2\pi)), if (\sin x = 0), then (x = 0,\pi).
  • Case 2: (2\cos x+\sqrt{3}=0) Solve for (\cos x): (\cos x=-\frac{\sqrt{3}}{2}). On the interval ([0,2\pi)), if (\cos x =-\frac{\sqrt{3}}{2}), then (x=\frac{5\pi}{6},\frac{7\pi}{6}).

Answer:

The solution set is (0,\frac{5\pi}{6},\pi,\frac{7\pi}{6})