Given $\cos(2x)=-\sin(x)$:

A. Solve the equation algebraically for the exact value of the solution(s) on the interval $[0,2\pi]$

B. Verify the answer(s) in part A using ZERO or INTERSECT features of your graphing calculator.

Any help is appreciated

  • 4
    $\begingroup$ Hint: $\cos (2x)=1-2\sin^2 x$. Use this and you have a quadratic equation in the variable $\sin x$. $\endgroup$ – David Mitra May 12 '14 at 22:17
  • 1
    $\begingroup$ Since you can't do much with the linear term $ \ -\sin x \ $ , write $ \ \cos 2x \ $ in the form $ \ 1 - 2 \sin^2 x \ $ . You will now have a quadratic equation in $ \ \sin x \ $ . $\endgroup$ – colormegone May 12 '14 at 22:17
  • $\begingroup$ And for the graphing calculator part, there are any number of videos like this one (youtube.com/watch?v=tIAtBsfgXxc) to help you, I'm sure. $\endgroup$ – grantfgates May 12 '14 at 22:20

Recall the trigonometric identity $\cos(2x)=1-2\sin^2x$. Now we can rewrite our equation in terms of sine. $$1-2\sin^2x=-\sin x$$ $$2\sin^2x-\sin x-1=0$$ Notice something? This is a quadratic equation in sine, which we can solve for. Let $\sin x=y$. $$2y^2-y-1=0$$ $$(y-1)(2y+1)=0$$ $$y=1, \ -\frac 12$$ Reverse the substitution. $$\sin x=1, \ -\frac 12$$ Break this up into two cases. $$\sin x=1$$ $$\sin x=-\frac 12$$ Let's solve $\sin x=1$ first. I know that one solution is $\frac{\pi}2$. This is also the only solution in the interval $[0, \ 2\pi]$ $$x=\frac{\pi}2$$ Now for $\sin x=-\frac 12$. Using the CAST rule, I know that one or more solutions will be between $\pi$ and $2\pi$. These solutions are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.

Therefore the solutions are: $$\color{green}{x=\frac{\pi}{2}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}}$$

To find the solutions graphically, first move $\cos(2x)$ to the right hand side. $$0=-\cos(2x)-\sin(x)$$ Replace $0$ with $y$. $$y=-\cos(2x)-\sin(x)$$ This is the equation of the graph, with the zeroes as the solutions. Here is what the graph looks like:

Graph of $y=-\cos(2x)-\sin(x)$

The zeroes within the interval $[0, \ 2\pi]$ are $\frac{\pi}{2},$ $\frac{7\pi}{6}$, and $\frac{11\pi}{6}$, which is the same as our solutions.

  • $\begingroup$ What's the app you use? Looks good. $\endgroup$ – Mike Miller May 13 '14 at 22:51
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    $\begingroup$ @BritMiller Search "Free Graphing Calculator" on the Apple Store. It should pop up as the first result. To make sure you find the right one, the icon is of a red coloured parabola with a black background. Enjoy! $\endgroup$ – TrueDefault May 13 '14 at 22:55
  • $\begingroup$ Quite a nice app, thanks! $\endgroup$ – Mike Miller May 14 '14 at 9:12

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