Question

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Find the solutions of each equation on the interval [0, 2π). Show work.

Answer 1

Answer: 0π

Explanation:

`sin\bigg((3\pi)/(2)+x\bigg)+sin\bigg((3\pi)/(2)+x\bigg)=-2\\\\\\2sin\bigg((3\pi)/(2)+x\bigg)=-2\\\\\\sin\bigg((3\pi)/(2)+x\bigg)=-1\\\\\\(3\pi)/(2)+x\bigg=sin^(-1)(-1)\\\\\\(3\pi)/(2)+x=(3\pi)/(2)\\\\\\\large\boxed{x=0}`