Find all solutions of the equation (sec(x))^2−2=0
The answer is A+Bk where k is any integer and 0<A<pi/2
Find:
A= pi/4?
B=??
done
http://www.jiskha.com/display.cgi?id=1350905280
To find all solutions of the equation (sec(x))^2 - 2 = 0, we first need to rewrite it in terms of trigonometric functions.
Let's start by expressing sec(x) in terms of cosine (cos(x)):
sec(x) = 1/cos(x)
Now, substitute sec(x) in the equation:
(1/cos(x))^2 - 2 = 0
To simplify further, multiply both sides of the equation by cos(x)^2:
1 - 2cos(x)^2 = 0
Rearrange the equation:
2cos(x)^2 - 1 = 0
Now, let's rewrite cos(x)^2 in terms of cosine 2x:
cos(x)^2 = (1 + cos(2x))/2
Substitute cos(x)^2 in the equation:
2(1 + cos(2x))/2 - 1 = 0
Simplify:
1 + cos(2x) - 1 = 0
Combine like terms:
cos(2x) = 0
Now we have:
cos(2x) = 0
To find all solutions, we need to find the values of 2x that satisfy cos(2x) = 0.
Recall that the solutions to this equation occur when the cosine function equals zero. The cosine function is zero at x = π/2 + πk, where k is any integer.
Since we have 2x, divide x = π/2 + πk by 2:
x = π/4 + πk/2
Now we have the general form of solutions: x = π/4 + πk/2, where k is any integer.
To find A and B, we need to observe the given conditions: 0 < A < π/2.
From the general solutions, we can see that when k = 0, x = π/4, which satisfies the condition 0 < x < π/2.
Therefore, A = π/4.
Regarding B, it is not explicitly mentioned in the question what B represents. B could be any real number associated with the variable k. Without further information, we cannot determine the specific value of B.
So, the answer to the equation is A + Bk, where k is any integer and 0 < A < π/2. In this case, A = π/4. The value of B will depend on the context of the problem.