Question

Can someone explain to me why sin(2ϴ/3)= -1 has no solution. The question is "Exercises 25-38 involve equations with multiple angles. Solve each equation on the interval [0,2pi). Please ASAP. Thank you! :)

Answer 1

In the interval $[0,2\pi)$, you have $\sin x=-1$ for $x=\dfrac{3\pi}2$. Replacing $x=\dfrac{2\theta}3$, you get

$\dfrac{2\theta}3=\dfrac{3\pi}2\implies\theta=\dfrac32\times\dfrac{3\pi}2=\dfrac{9\pi}4$

But $2\pi=\dfrac{8\pi}4<\dfrac{9\pi}4$, so that solution for $\theta$ falls outside the interval $[0,2\pi)$.