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I was doing a simple trig question when it turned out I was missing several answers. I have read somewhere that it is possible to lose information about the signs when dealing with squares and square roots and wondered if something similar happened here.

The question is to find all values of x which satisfy $\sin 4x=\sin 2x$, where $x \in [0, \pi]$

This is what I did: $$2\sin 2x\cos 2x=2\sin x\cos x$$ $$2(2\sin x\cos x)(\cos^2x-\sin^2x)=2\sin x\cos x$$ $$2\cos^2x-2\sin^2x=1$$ $$2\cos^2x-2(1-\cos^2x)=1$$ $$4\cos^2x-3=0$$ $$\cos x={\pm\sqrt3\over2}$$ $$x={\pi\over6},{5\pi\over6},{7\pi\over6},{11\pi\over6}$$ Apparently I am missing $x={0},{\pi\over2},{\pi},{3\pi\over2}, 2\pi$. Does anyone know if I did something in my solution which lead to the loss of over half the answers? Likewise, how could I do it differently to retain those answers?

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Certainly canceling $\sin x \cos x$ on the third line is dangerous if $x$ should make that quantity zero. –  Dylan Moreland Dec 18 '11 at 8:12

2 Answers 2

up vote 3 down vote accepted

You divided by $\sin x\cos x$ in the third step. You can't divide by 0, so you need to examine what happens when $\sin x=0$ or $\cos x=0$.

In this case, when $\sin x=0$ or $\cos x=0$, you have solutions to the equation.

Just look at $$\tag{1}2(2\sin x\cos x)(\cos^2x-\sin^2x)=2\sin x\cos x.$$ If $\sin x=0$ or $\cos x=0$, the above equation becomes $0=0$; and the other solutions can be found by solving the equations $\sin x=0$ and $\cos x=0$.

If both $\sin x\ne 0$ and $\cos x\ne0$, you may cancel them on both sides, and equation (1) is equivalent to $$2\cos^2x-2\sin^2x=1.$$

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You can also use the $\sin(C) - \sin(D)$ formula. So your equation becomes,

\begin{align*} \sin(4x) - \sin(2x) &= 0 \\ \Rightarrow 2 \cos\biggl(\frac{4x+2x}{2}\biggr) \cdot \sin\biggl(\frac{4x-2x}{2}\biggr) &= 0 \\ \Rightarrow 2 \cos(3x) \cdot\sin(x) =0 \end{align*}

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