Question

Trigonometry Question.

Answer 1

Answer:

There are two solutions: [format = (x,y)]

1. (π/2,0) 2. (π/2,π)

Step-by-step explanation:

$2 { (\sin(x)) }^{2} = {2}^{ {( \sec(y)) }^{2} }$

$= > {( \sin(x) )}^{2} = {2}^{ {( \sec(y) )}^{2} } \div 2 = {2}^{ {( \sec(y) )}^{2} - 1 }$

Now, in the LHS, we have sin²x.

sin²x can take values from 0 to 1.

Hence 0 ≤ LHS ≤ 1 .................1

In the RHS, we have sec²x.

sec(x) is greater than 1 or less than -1. Hence sec²x is always greater than 1 . Hence

sec²x - 1 ≥ 0

This means that in RHS, 2 has a positive power.

Hence RHS ≥ 1 ...................2

From 1 and 2, we get LHS = RHS = 1

From 1:

sin²x = 1

=> sin(x) = ± 1

But is domain (0,π), sine function is positive.

hence sin x = 1

=> x = π/2

From 2:

2^(sec²y-1) = 1

=> sec²y-1 = 0 => sec²y = 1

=> sec y =± 1 => cos y =± 1

=> cos y = 1 or cos y = -1

=> y = 0 or y = π

So ordered pairs are :

(π/2,0) and (π/2,π)