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My question is:
How do I find all solutions in the interval $[0, 2\pi)$ of the equation $\cos 2x - \sin x = 1$?

Any pointers into the direction I should be taking would be very helpful.
Thank you in advance.

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Hint: Is there any way to rewrite $\cos(2x)$ in terms of $\sin(x)$? –  user5137 Sep 13 '11 at 19:59
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2 Answers 2

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The double-angle formula gives that $$\cos(2x) = \cos(x+x) = \cos^2(x) - \sin^2(x) = (1-\sin^2 x)-\sin^2x = 1-2\sin^2 x.$$ Substituting into $$\cos(2x) - \sin x = 1$$ should make it more tractable.

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So If I substitute, I get $1 - 2sin^2x - sinx = 1$, which is $-2sin^2x - sinx = 0$. Is there a way to simplify this further? –  Mike Gates Sep 13 '11 at 20:06
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@Mike Gates: It's not complete. From $-2\sin^2 x - \sin x = 0$ you get $-\sin(x)\Bigl( 2\sin x + 1\Bigr)=0$. For the product to be zero, you need either $\sin(x)=0$ (which is what you already solved), or $2\sin(x)+1=0$, which you have not dealt with yet. (For Srivatsan's suggestion: if you replace $\sin x$ with $y$, you get $-2y^2 - y = 0$, or $2y^2+y = 0$. That's a quadratic equation, and you probably know how to find all solutions to that; for each solution $y$, you then need to solve $\sin x = y$ for $x$). –  Arturo Magidin Sep 13 '11 at 20:22
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@Mike: If you had followed the advice in my answer, you would have noticed on your own that $0$ and $\pi$ aren't the only solutions... ;-) –  Hans Lundmark Sep 13 '11 at 20:24
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I see that theres one more solution just beyond $\pi$ when I graphed the 2 functions, but I still can't reason out what it is. I tried following @Arturo's previous comment, but to no avail. –  Mike Gates Sep 13 '11 at 20:29
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@Mike: What's the problem? You want to solve $2\sin x + 1 =0$, or $\sin x = -\frac{1}{2}$. Are there any values of $x$ on $[0,2\pi)$ where $\sin(x)$ takes the value $-\frac{1}{2}$? Yes! There's two of them. So those two values of $x$ are also solutions. –  Arturo Magidin Sep 13 '11 at 20:31
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Arturo's hint already tells you how to proceed with the computations. To check whether the solutions that you calculate seem plausible, and to make sure that you haven't missed any solutions, it's useful to write the equation as $$\cos 2x = \sin x + 1.$$ Then you can easily plot the curves $y=\cos 2x$ (which looks like the familiar $y=\cos x$ except that it oscillates twice as fast) and $y=\sin x+1$ (the familiar sine curve, but moved one unit upwards), and see roughly where they intersect (which they do at the $x$-values that satisfy the equation).

(It should look like this, but try it yourself before you peek!)

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Is it not just $0$ and $\pi$? –  Mike Gates Sep 13 '11 at 20:26
    
@Mike: Yes, it is not just $0$ and $\pi$; there are other solutions. –  Arturo Magidin Sep 13 '11 at 20:29
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@Mike: It's a bit tricky to see in the first Wolfram plot, since it doesn't go all the way to $2\pi$, but there are some solutions in the interval $[-\pi,0]$ that will have counterparts one period later, in the interval $[\pi,2\pi]$. Look at height $y=0.5$, and you will hopefully find them! –  Hans Lundmark Sep 13 '11 at 20:31
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@Hans Lundmark: I strongly agree with pointing beyond the formal manipulational aspect of the problem. –  André Nicolas Sep 13 '11 at 21:06
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