# Solving $\cos(2x)=-\sin(x)$

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

• Hint: $\cos (2x)=1-2\sin^2 x$. Use this and you have a quadratic equation in the variable $\sin x$. – David Mitra May 12 '14 at 22:17
• 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 \$ . – colormegone May 12 '14 at 22:17
• 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. – 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:
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.