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I have been trying to solve this question:

$$\sin \frac12x =-3 \cos 2x$$

However, I do not even know where to begin. I have graphed it, so I know what x must be and that there are 4 solutions, but I just do not know the equations.

What should I do to solve it?

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    $\begingroup$ Half and double angle formulas are a good start. $\endgroup$
    – Ennar
    Commented Apr 17, 2018 at 22:41
  • $\begingroup$ $\cos 2x = \cos^2 x - \sin^2 x \implies \cos 2x = (\cos^2 \frac 12 x - \sin^2 \frac 12 x)^2 - (2\sin \frac12 x\cos \frac 12x)^2$ simpify to get a quartic equation in terms of $\sin \frac12 x$ $\endgroup$
    – Doug M
    Commented Apr 17, 2018 at 22:53
  • $\begingroup$ @DougM The problem is that you get a quartic ($24s^4-24s^2+s+3=0$) in $s=\sin \frac12 x$ that has 4 irrational real roots (and the irrationals are not the "famous" trig ratios). Using the general quartic solution is hard and not very revealing in this case. The only reasonable approach is a numerical one. Not really satisfying. But if it is an approach the OP is willing to consider, try this: 1728.org/quartic.htm $\endgroup$
    – Deepak
    Commented Apr 17, 2018 at 23:22
  • $\begingroup$ @Deepak I came to that same equation and realization myself. $\endgroup$
    – Doug M
    Commented Apr 17, 2018 at 23:34

2 Answers 2

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$$\sin\left(\frac{x}{2}\right) = \sqrt[]{\frac{1-\cos(x)}{2}} = -3\cos(2x)$$ $$\frac{1-\cos(x)}{2} = 9\cos^2(2x) \rightarrow 0= 18\cos^2(2x)+\cos(x)-1$$

From here it is a lot of bashing/algebra knowing that $\cos(2x) = \sin^2(x)-\cos^2(x)$

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  • $\begingroup$ +1 use $\cos(2x)=2\cos^2x-1$ instead $\endgroup$
    – user577215664
    Commented Apr 17, 2018 at 23:07
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Let $\sin\dfrac x2=a,\cos x=1-2a^2$

$$\implies a=-3\left(2(1-2a^2)^2-1\right)$$

$$24a^4-24a^2+a+3=0$$

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