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No idea how to solve this: Find all solutions if 0 ≤ x < 2π. Use exact values only. (Enter your answers as a comma-separated list.) sin x cos 2x + cos x sin 2x = 1/2
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\[\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\]
that's a hint
3x=1/2 but where do i go from there
you mean sin(3x)=1/2
yes
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\[0 \le x < 2\pi \\ \text{ let } u=3x \\ \text{ so } \frac{u}{3}=x \\ 0 \le x < 2\pi \text{ implies } 0 \le \frac{u}{3} < 2\pi \\ \text{ so solve } \sin(u)=\frac{1}{2} \text{ on } 0 \le u < 6 \pi \\ \text{ then after you are done replace u with 3x and solve for x}\]
i have no idea what you are saying
sin(u)=1/2 can be solved on [0,6pi) using the unit circle
ohhhh
soooo
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