Question

find the value of A .if i.(cosec 2A-2)(cot 3A-1)=0 ii.cos 3A(2sin2A-1)=0

Answer 1

$$\mathsf {I}. \big (c o s e c (2 A - 2) \big) (\cot (3 A - 1)) = 0.$$

Then

$$c o s e c (2 A - 2) = 0 \text{ or } \cot (3 A - 1) = 0.$$

As $\text{cosec}(2A - 2) = \frac{1}{\sin(2A - 2)} \neq 0$, so $\cot(3A - 1) = 0$.

$$\cot (3 A - 1) = \frac {\cos (3 A - 1)}{\sin (3 A - 1)} = 0.$$

Then $\cos (3A - 1) = 0$, $3A - 1 = \frac{\pi}{2} + \pi n$, $n \in \mathbb{Z}$.

Finally, $A = \frac{1}{3} + \frac{\pi}{6} + \frac{\pi n}{3}$, $n \in \mathbb{Z}$.

II. $\cos (3A) * (2 \sin (2A) - 1) = 0$

Then

$$\cos (3 A) = 0 \text{ or } (2 \sin (2 A) - 1) = 0.$$

For the first equation:

$$3 A = \frac {\pi}{2} + \pi n, n \in \mathbb {Z};$$ $$A = \frac {\pi}{6} + \frac {\pi n}{3}, n \in \mathbb {Z}.$$

For the second:

$$\sin (2 A) = \frac {1}{2},$$ $$2 A = \frac {\pi}{6} + 2 \pi k \text{ or } 2 A = \frac {5 \pi}{6} + 2 \pi k, k \in \mathbb {Z};$$ $$A = \frac {\pi}{12} + \pi k \text{ or } A = \frac {5 \pi}{12} + \pi k, k \in \mathbb {Z}.$$

Finally, the answer is: $A = \frac{\pi}{6} + \frac{\pi k}{3}$, $A = \frac{\pi}{12} + \pi k$ and $A = \frac{5\pi}{12} + \pi k$ for all $k \in \mathbb{Z}$.