Okay, so the options that have the red dots are my answers. I am particularly uncertain about number 18. Please could you see if you agree with me.
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$\begingroup$Your solutions for questions 17, 19, and 20 are correct. As for question 18: \begin{align*} \sin(3x) + \cos(3x) & = 0\\ \sin(3x) & = -\cos(3x)\\ \tan(3x) & = -1\\ 3x & = -\frac{\pi}{4} + k\pi, k \in \mathbb{Z}\\ x & = -\frac{\pi}{12} + \frac{k\pi}{3} \end{align*} Since $-\pi < x < \pi$, $-2 \leq k \leq 3$. Thus, the solution set is $$\left\{-\frac{3\pi}{4}, -\frac{5\pi}{12}, -\frac{\pi}{12}, \frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12}\right\}$$
$\endgroup$ $\begingroup$It's $$\sin(3x+45^{\circ})=0$$ or $$3x=-45^{\circ}+180^{\circ}k$$ or $$x=-15^{\circ}+60^{\circ}k,$$ where $k\in\mathbb Z$.
Thus, $$-180^{\circ}<-15^{\circ}+60^{\circ}k<180^{\circ}$$ or $$-2.75<x<3.25,$$ which gives $k\in\{-2,-1,0,1,2,3\}$ and the 3) is valid.
Done!
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