Pre-Calculus: Solve: square root 2 sin D cos D = sin D

Solve: square root 2 sin D cos D = sin D; D E [-180,180]
I think you mean \(\displaystyle D\in (-\pi,\pi)\) an open interval.

Does \(\displaystyle D=0\) solve it? If \(\displaystyle D\ne 0\) can you say that \(\displaystyle \cos(D)=\dfrac{\sqrt2}{2}.\)
 
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