# 3 dB bandwidth of a filter

I am solving one of the question from signal processing about calculating the 3dB bandwidth of the filter. But I am stuck once I got the magnitude response of the filter. Can someone help me with this?

From the given transfer function you can see that the filter has a low pass character. The gain at DC is given by $$H(1)=1$$. This is also the maximum value of $$|H(e^{j\omega})|$$. So you can try to solve
$$\big|H(e^{j\omega_c})\big|^2=\frac{\max_{\omega}|H(e^{j\omega})\big|}{2}=\frac12\tag{1}$$
because this equation defines the frequency $$\omega_c$$ at which the filter gain equals $$1/\sqrt{2}$$, which is the definition of the $$3$$ dB cut-off frequency.
Your calculation of the magnitude of $$H(e^{j\omega})$$ is correct, so you can just continue by solving $$(1)$$.
As a final note, there are values of $$a$$ for which the $$3$$ dB cut-off frequency doesn't exist because the filter never reaches an attenuation of $$3$$ dB. You can try to find that limit of $$a$$ as a bonus exercise.