Full solution set of $\sin2\theta=-1$

Question

I have the equation $\sin2\theta=-1$. I am supposed to solve for $\theta$, which I did: $$\frac{3\pi}4+2\pi k\quad k\in\mathbb Z$$ However, the actual answer is: $$\frac{3\pi}4+\pi k\quad k\in\mathbb Z$$ How come they are only adding $\pi k$ instead of $2\pi k$? I thought $2\pi$ was a full rotation to find any additional related angles?

Answer 1

$\sin2\theta=-1$ gives $\displaystyle 2\theta=2\pi k+\frac{3\pi}{2}$. So, $\displaystyle \theta=\pi k+\frac{3\pi}{4}$.

You have to find the general solution for $2\theta$ and use it to find the general $\theta$.

Answer 2

$$\sin 2 \theta = -1$$

$$2\theta = \frac{3(2\pi)}{4} + 2k\pi, k \in \mathbb{Z}$$

Dividing by $2$ gives

$$\theta = \frac{3(\pi)}{4} + k\pi, k \in \mathbb{Z}$$

Answer 3

If you define $\alpha = 2\theta$, then $\sin \alpha = -1$ implies $\alpha = \dfrac{3\pi}{2}+2\pi k$ as you suggested. But notice that here we have $$\alpha = 2\theta = \dfrac{3\pi}{2}+2\pi k\implies \theta = \dfrac{3\pi}{4}+\pi k$$

Answer 4

$$0=1+\sin2\theta=(\cos\theta+\sin\theta)^2$$

$$\implies\cos\theta+\sin\theta=0\iff\tan\theta=-1=-\tan\dfrac\pi4=\tan\left(-\dfrac\pi4\right)$$

$$\implies\theta=m\pi-\dfrac\pi4$$ where $m$ is any integer

Answer 5

One more:

Set $\alpha =2\theta.$

Find solutions to $\sin\alpha =-1.$

One solution is $(3π)/2.$

Since $\sin$ has period $2π$ ,

we can add $k2π$ , $k \in \mathbb{Z}$:

$\alpha = (3π)/2 +k2π,$

or in terms of $\theta:$

$2\theta = (3π)/2 + k2π,$

and finally:

$\theta = (3π)/4 + kπ$, $k \in.\mathbb{Z}.$