Asked by math
                Find all solutions of the equation 
sec^2x-2=0.
The answer is A+Bkπ where is any integer and 0<a<π/2
            
            
        sec^2x-2=0.
The answer is A+Bkπ where is any integer and 0<a<π/2
Answers
                    Answered by
            Bosnian
            
    sec² ( x ) - 2 = 0 
Add 2 to both sides
sec² ( x ) = 2
sec( x ) = ± √ 2
sec( x ) = √ 2 and sec( x ) = - √ 2
The solutions are:
x = sec⁻¹ ( √ 2 )
x = π / 4 , x = 7 π / 4
and
x = sec⁻¹ ( - √ 2 )⁻
x = 3 π / 4 , x = 5 π / 4
The period of sec ( x ) is 2 k π so the solutions are:
x = π / 4 + 2 k π , x = 7 π / 4 + 2 k π
and
x = 3 π / 4 + 2 k π , x = 5 π / 4
Condition:
x = A + B k π
0 < A < π / 2
Solution is x = π / 4 + 2 k π becouse:
3 π / 4 > π / 2
5 π / 4 > π / 2
7 π / 4 > π / 2
Solution:
x = π / 4 + 2 k π
    
Add 2 to both sides
sec² ( x ) = 2
sec( x ) = ± √ 2
sec( x ) = √ 2 and sec( x ) = - √ 2
The solutions are:
x = sec⁻¹ ( √ 2 )
x = π / 4 , x = 7 π / 4
and
x = sec⁻¹ ( - √ 2 )⁻
x = 3 π / 4 , x = 5 π / 4
The period of sec ( x ) is 2 k π so the solutions are:
x = π / 4 + 2 k π , x = 7 π / 4 + 2 k π
and
x = 3 π / 4 + 2 k π , x = 5 π / 4
Condition:
x = A + B k π
0 < A < π / 2
Solution is x = π / 4 + 2 k π becouse:
3 π / 4 > π / 2
5 π / 4 > π / 2
7 π / 4 > π / 2
Solution:
x = π / 4 + 2 k π
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