Solve, $\sin(2x) = \cos(x)$, on the interval $[0, 2\pi)$.

Question

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.

Answer 1

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 $x \in \{\pi/2, \,3\pi/2, \, \pi/6, \, 5\pi/6\}$.

Answer 2

$$ \sin 2x = 2 \sin x \cos x$$

Therefore the equation is equivalent to $$ 2\sin x \cos x= \cos x$$

You can factor $ \cos x$ and continue from there.

Answer 3

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.