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Solve Cos(2 theta) = -1/2
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\(\bf cos(2\theta)=-\cfrac{1}{2}\implies 2cos^2(\theta)-1=-\cfrac{1}{2}\) and simplify it like the previous one, and take \(\Large cos^{-1}\) to get \(\theta\) like in the previous one
thank you can you answer another question? Sin(3theta)=-1?
hmmm
well... for one I need to dash in a few secs but the triple angle identity wouild be \(\bf {\color{brown}{ sin(3\theta)\to 3sin(\theta)-4sin^3(\theta)}}\) though I didn't find much headway there so I'd instead split it in a 2 angle sum or \(\bf sin(3\theta)=-1\qquad \begin{cases} sin(\theta+2\theta)\\ sin(\theta)cos(2\theta)+cos(2\theta)sin(\theta)\\ sin(\theta)[1-2sin^2(\theta)]+cos(\theta)[2sin(\theta)cos(\theta)] \end{cases}\)
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