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

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

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