Find all solutions of the equation
sec^2x-2=0.
The answer is A+Bkπ where is any integer and 0<a<π/2
Well, well, what do we have here? An equation with sec^2x? Are you sure you're not just trying to clown around with me? Alright, let's solve this.
To find the solutions for the equation sec^2x - 2 = 0, we need to isolate sec^2x by itself. So let's add 2 to both sides:
sec^2x = 2
Now, to get rid of that pesky square, we need to take the square root of both sides:
√(sec^2x) = ±√2
Remember, since sec^2x is positive, we can ignore the negative square root. So we're left with:
sec x = √2
Now, we can use the definition of secant to rewrite this as:
cos x = 1/√2
To find the solutions within the given range of 0 < a < π/2, we need to think about the unit circle and the values of cosine in that range. The unit circle tells us that cosine is positive in the first and fourth quadrants.
So, in the first quadrant (0 to π/2), the value of x that satisfies the equation cos x = 1/√2 is x = π/4.
Now, in the fourth quadrant (3π/2 to 2π), the value of x that satisfies the equation cos x = 1/√2 is x = 7π/4.
Remember, the general formula for finding all the solutions to this type of equation is A + Bkπ, where A is any particular solution and B is an integer.
So, in this case, the solutions to sec^2x - 2 = 0 within the given range of 0 < a < π/2 are:
π/4 + kπ, where k is an integer.
Hope that clears things up and adds a touch of clownishness to your day!
To find all solutions of the equation sec^2x - 2 = 0, we can start by isolating sec^2x:
sec^2x = 2.
Next, we take the square root of both sides to solve for sec x:
sec x = √2.
Now, remember that sec x is equal to 1/cos x. So we can rewrite the equation as:
1/cos x = √2.
Taking the reciprocal of both sides gives us:
cos x = 1/√2.
To solve for x, we need to find the values of x where the cosine function equals 1/√2.
Using the unit circle or a trigonometric table, we know that cos(π/4) = √2/2.
So, we can write:
cos x = cos(π/4).
To find all solutions, we can use the periodicity of the cosine function. The cosine function has a period of 2π, so we add integer multiples of 2π to find other solutions.
Thus, the solutions are:
x = π/4 + 2πk,
where k is any integer.
Since we want 0 < a < π/2, the solution is:
x = π/4 + 2πk, where k is any integer.
Therefore, the answer is A + Bkπ, where A = π/4 and B = 2.
To find all the solutions of the equation sec^2x-2=0, we can follow these steps:
Step 1: Solve the equation for sec^2x: sec^2x = 2.
Step 2: Take the square root of both sides: secx = ±√(2).
Step 3: Find the values of x that satisfy the equation secx = √(2) and secx = -√(2). These will be the values that make the equation true.
Step 4: To do this, we need to remember the unit circle and the definitions of the trigonometric functions. Also, we know that secx = 1/cosx, so we can rewrite the equation as follows:
1/cosx = ±√(2).
Step 5: Simplify the equation to cosx = ±1/√(2).
Step 6: Rationalize the denominator by multiplying the numerator and denominator by √(2):
cosx = ±√(2)/2.
Step 7: Determine the values of x that satisfy cosx = √(2)/2 and cosx = -√(2)/2. These values will correspond to points on the unit circle where the x-coordinate is √(2)/2 or -√(2)/2.
Step 8: On the unit circle, we see that the values of x where cosx = √(2)/2 are x = π/4 and x = 7π/4.
Step 9: Similarly, the values of x where cosx = -√(2)/2 are x = 3π/4 and x = 5π/4.
Step 10: Combining these solutions, we have x = π/4, x = 3π/4, x = 5π/4, and x = 7π/4.
Step 11: Expressing the solutions as A + Bkπ, where A is between 0 and π/2, and k is any integer, we get the final solutions: x = π/4 + kπ, x = 3π/4 + kπ, x = 5π/4 + kπ, and x = 7π/4 + kπ, where k is any integer and 0 < π/4 < π/2.
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