Which of these is a possible solution for sin^2 x-1/2 =0 in the interval x ∈ [0,2pi]?
a. x = pi/4
b. x = 7pi/4
c. x = 5pi/4
d. all of the above
sin^2x = 1/2
sinx = ±1/√2
so you now have the reference angle of π/4, in all four quadrants.
xcv
To solve the equation sin^2(x) - 1/2 = 0, we can rearrange it as sin^2(x) = 1/2.
Taking the square root of both sides, we have sin(x) = ±√(1/2).
In the interval x ∈ [0,2π], the solutions for sin(x) = √(1/2) are x = π/4 and x = 3π/4.
Similarly, the solutions for sin(x) = -√(1/2) are x = 5π/4 and x = 7π/4.
Therefore, the correct answer is d. all of the above.
To find the possible solutions for the equation sin^2(x) - 1/2 = 0 in the interval x ∈ [0,2π], we can use the following steps:
Step 1: Rewrite the equation in terms of sin(x):
sin^2(x) - 1/2 = 0
Step 2: Add 1/2 to both sides of the equation:
sin^2(x) = 1/2
Step 3: Take the square root of both sides of the equation:
sin(x) = ±√(1/2)
Step 4: Determine the possible values for sin(x):
sin(x) = ±√(1/2) can be written as:
sin(x) = ± (√2/2)
The values of sin(x) that equal (√2/2) are known from the unit circle. In the interval x ∈ [0,2π], the values of x where sin(x) = (√2/2) are π/4 and 3π/4.
Similarly, the values of sin(x) that equal -(√2/2) are also known from the unit circle. In the interval x ∈ [0,2π], the values of x where sin(x) = -(√2/2) are 5π/4 and 7π/4.
Therefore, the possible solutions for sin^2(x) - 1/2 = 0 in the interval x ∈ [0,2π] are:
a. x = π/4
b. x = 7π/4
c. x = 5π/4
So, the correct answer is d. all of the above.