The second solution of $\sin\left(x\right)=\cos\left(2x\right)$

Question

I'm self-learning trigonometry, and I came across this equation: $$\sin\left(x\right)=\cos\left(2x\right)$$ First I transformed $\sin\left(x\right)$ into $\cos\left(\frac{\pi }{2}-x\right)$. Then I took the $\arccos$ of both sides and solved the equation. The solution I arrived at was $$x=\frac{\pi }{6}+\frac{2\pi n}{3},\:n \in \mathbb{Z}$$ However, there is a second solution, namely $$-\frac{\pi }{3}+2\pi n,\:n\in \mathbb{Z}$$ Thing is: I can't comprehend why this is a second solution and how could I have arrived at it.

Answer 1

Guide:

$$\sin(x) = \cos(2x) = 1-2\sin^2(x)$$

Let $u=\sin(x)$, we have

$$u=1-2u^2$$ $$2u^2+u-1=0$$

Try to solve this quadratic equation.

Answer 2

In equation $\sin x=\sin y$ you have one solution $x=y+2k\pi$ with $k\in \mathbb Z$. However you can also write this $$\sin y = -\sin(-y)=-\sin(\pi+y)=\sin(-y-\pi)$$ This yields $x=-y-\pi+2k\pi$

Answer 3

Within a given copy of the circle (meaning a half-open interval of length $2\pi$), cosine and sine are two-to-one functions except at their maxima/minima. The two solutions to $\cos(\theta)=x$ lie on the same vertical line, the two solutions to $\sin(\theta)=y$ lie on the same horizontal line.

That means that when you write $\cos(\pi/2-x)=\cos(2x)$, even if you assume that $\pi/2-x$ and $2x$ are in the same copy of the circle, you still have to worry about the other solution within that same copy of the circle. One way to write this algebraically is $\pi/2-x=\pm 2x+2 \pi n$, taking advantage of the property $\cos(-x)=\cos(x)$ (which is just this "vertical line" idea). For $\sin$ you instead would use $\sin(\pi-x)=\sin(x)$ (which is just this "horizontal line" idea).

Answer 4

We can use that

$$\sin\alpha =\cos\beta \iff \beta=\frac{\pi}2-\alpha+2k\pi \quad \lor \quad \beta=-\frac{\pi}2+\alpha+2k\pi\quad k\in\mathbb{Z}$$

that is

$$\sin\left(x\right)=\cos\left(2x\right) \iff 2x=\frac{\pi}2-x+2k\pi \quad \lor \quad 2x=-\frac{\pi}2+x+2k\pi \quad k\in\mathbb{Z}$$

and therefore the solutions are

$$x= \frac{\pi}6+\frac23k\pi \quad k\in\mathbb{Z}$$

Note that $x=-\frac{\pi }{3}$ is not a solution indeed

$$\sin\left(-\frac{\pi }{3}\right)\neq \cos\left(-\frac{2\pi }{3}\right)$$