I need to solve, $\sin(2x) = \cos(x)$, on the interval $[0, 2\pi)$.
The answer is $x = \{ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2}, \frac{3\pi}{2} \}$
I'm lost. I don't know how to get radian/constant answers from variables like $x$.
This is part of a handout, and on it, professor wrote, "Note, in most of these problems, your first step is usually a front/backdoor identity."
I know about trig identities, but I don't know what a "front/backdoor" identity means.
It seems like I might be able to do something with $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, but I'm not sure what.
$\endgroup$3 Answers
$\begingroup$Observe that $\sin(2x)=2\sin x \cos x$, so that$$ \sin(2x) = \cos x \quad \iff \quad \cos x(2\sin x-1) = 0 \quad \iff \quad \cos x = 0 \;\text{ or } \; 2\sin x-1=0. $$
The final pair of equations is solved in a standard way. The equation $\cos x = 0$ has two solutions in $[0,2\pi)$, namely $x = \pi/2, \,3\pi/2$. Furthermore,$$ 2\sin x -1 = 0 \quad \iff \quad \sin x = 1/2,$$which also has two solutions in $[0,2\pi)$, namely $x = \pi/6, \, 5\pi/6$.
Hence the solutions to the original equation are$x \in \{\pi/2, \,3\pi/2, \, \pi/6, \, 5\pi/6\}$.
$\endgroup$ 10 $\begingroup$$$ \sin 2x = 2 \sin x \cos x$$
Therefore the equation is equivalent to $$ 2\sin x \cos x= \cos x$$
You can factor $ \cos x$ and continue from there.
$\endgroup$ 2 $\begingroup$so we have:$$\sin(2x)=\cos(x)$$note from the double angle formula that:$$\sin(2x)=2\sin(x)\cos(x)$$so we get:$$2\sin(x)\cos(x)=\cos(x)$$from this we get two types of solutions:$$\cos(x)=0\tag{1}$$$$2\sin(x)=\cos(x)\tag{2}$$if we continue first with $(1)$we get:$$\cos(x)=0$$$$x=\frac\pi2,\frac{3\pi}2$$as we are restricted with $0\le x\le2\pi$. Now we can move on with $(2)$:$$2\sin(x)=\cos(x)$$rearranging gives:$$\tan(x)=\frac{1}{2}$$and there are a further two solutions to this.
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