In the figure given below, P and Q are centres of two circles, intersecting at B and C,

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part  of the circle.

Here, arc AEB subtends ∠APB at the centre and ∠ACB at C on the circle.

∴ ∠APB = 2∠ACB

$\Rightarrow \angle A C B=\frac{150^{\circ}}{2}=75^{\circ}$          ...(1)

Since ACD is a straight line, ∠ACB + ∠BCD = 180∘
⇒ ∠BCD = 180∘ − 75∘
⇒ ∠BCD = 105∘            ...(2)

Also, arc BFD subtends reflex ∠BQD at the centre and ∠BCD at C on the circle.

∴ reflex ∠BQD = 2∠BCD

$\Rightarrow$ reflex $\angle B Q D=2\left(105^{\circ}\right)=210^{\circ}$      ...(3)

Now,
reflex ∠BQD + ∠BQD = 360∘
⇒ 210∘ + x = 360∘
⇒ x = 360∘ − 210∘
⇒ x = 150∘

Hence, x = 150∘.