Question

csc(6a+ð/8)=sec(2a-ð/8)
Find values of A.

Answer 1

Since the original question was not posted in Western encoding, the strange character would reasonably be interpreted as π. So the question would be:

<i>Solve csc(6a+π/8) = sec(2a-π/8)</i>......(1)

Knowing that csc(x)=1/sin(x) and sec(x)=1/cos(x), we proceed to transform the equation algebraically:
1/sin(6a+π/8) = 1/cos(2a-π/8)
if sin(6a+π/8)≠0 and cos(2a-π/8)≠0, we can cross multiply to get:
cos(2a-π/8) = sin(6a+π/8) ......(2)

Using the identity
sin(θ)=cos(θ-π/2) .....(3)
we transform the right-hand-side of (2) into a cosine function.
cos(2a-π/8) = cos(6a+π/8-π/2)
cos(2a-π/8) = cos(6a-3π/8) ..... (4)

The general solution to (4) is:
±(2a-π/8) = 6a-3π/8 + 2kπ where k∈ℤ (integers)
The ± sign is applicable because cos(x)=cos(-x) for all x.

Case 1:
+(2a-π/8) = 6a-3π/8 + 2kπ
solve for a to get
a=π/16+kπ/2 ......(5)

Case 2:
-(2a-π/8) = 6a-3π/8 + 2kπ
solve for a to get
a=π/16+kπ/4 ......(6)

Since the solution (5) is a subset of (6), it would be tempting to declare the answer as given by (6).

Not so soon!

Remember we said that:
"if sin(6a+π/8)≠0 and cos(2a-π/8)≠0, we can cross multiply ...",
Now it is the time to check if those conditions are met!

By evaluating successively the solutions given by case 1 and case 2,
we will find that certain solutions of case 2 will result in:
sin(6a+π/8)=0 and cos(2a-π/8)=0. These are exactly the differences between case 2 and case 1. So we will retain case 1 as our only answer, namely:

<b>a=π/16+π/4, k∈ℤ (integers)</b>

Check my calculations and my thinking.

Answer 2

<b>a = π/16 + kπ/4, k∈ℤ</b> (integers)