Answer:
[tex]\displaystyle{\theta=\dfrac{5\pi}{4}\, , \, \dfrac{9\pi}{4}}[/tex]
Step-by-step explanation:
Given the equation [tex]\displaystyle{2\cos \theta = 2\sin \theta}[/tex]. We have to divide both sides by 2 which gives us [tex]\displaystyle{\cos \theta = \sin \theta}[/tex]. Then divide both sides by [tex]\displaystyle \cos \theta[/tex] which gives us [tex]\displaystyle{1=\dfrac{\sin \theta}{\cos \theta}}[/tex].
We know that:
[tex]\displaystyle{\dfrac{\sin \theta}{\cos \theta} = \tan \theta}[/tex]
So we can rewrite the equation as [tex]\displaystyle{1=\tan \theta}[/tex].
We know that [tex]\displaystyle{\tan \theta = 1}[/tex] when [tex]\displaystyle{\theta = \dfrac{\pi}{4}+\pi k}[/tex] for [tex]\displaystyle{k \in I}[/tex] (k is any integer). However, the equation is given with the interval [tex]\displaystyle{\pi \leq \theta \leq 3\pi}[/tex]. Therefore, we have to substitute k-values that satisfy the interval.
It appears that only k = 1, 2 which gives us [tex]\displaystyle{\theta_1 = \dfrac{\pi}{4}+\pi}[/tex] and [tex]\displaystyle{\theta_2 = \dfrac{\pi}{4}+2\pi}[/tex]can be used since both satisfies both values and interval.
Simplifying both solutions:
[tex]\displaystyle{\theta_1 = \dfrac{\pi}{4}+\dfrac{4\pi}{4} \, , \, \theta_2=\dfrac{\pi}{4}+\dfrac{8\pi}{4}}\\\\\displaystyle{\theta=\dfrac{5\pi}{4}\, , \, \dfrac{9\pi}{4}}[/tex]
