I am trying to do this problem and I only get so far. $\displaystyle 2 sin3x+1=0$ Thanks in advance.
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So what z solves? $\displaystyle 2 z +1=0 $ Then what y solves $\displaystyle sin y = z $ Then what x solves $\displaystyle 3x = y $
Originally Posted by Godzilla I am trying to do this problem and I only get so far. $\displaystyle 2 sin3x+1=0$ Thanks in advance. hi all
Put $\displaystyle sin(3x)=y$ and solve for $\displaystyle y$,the equation $\displaystyle 2y+1=0$. Remember,$\displaystyle sin(-\frac{\pi }{6})=-\frac{1}{2}$.
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