46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.
Give exact answers where possible.
e) csc2 x = 2 + csc x
To solve the equation csc^2(x) = 2 + csc(x), we can substitute cosec(x) as 1/sin(x):
(1/sin(x))^2 = 2 + 1/sin(x)
Now, let's simplify the equation by multiplying both sides by sin^2(x):
1 = (2sin^2(x)) + sin(x)
Next, let's rearrange the equation:
2sin^2(x) + sin(x) - 1 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, we will use factoring:
(2sin(x) + 1)(sin(x) - 1) = 0
Setting each factor equal to zero gives us two separate equations:
2sin(x) + 1 = 0 and sin(x) - 1 = 0
Solving the first equation:
2sin(x) + 1 = 0
2sin(x) = -1
sin(x) = -1/2
Using the unit circle or a reference angle, we find that x = 7π/6 and x = 11π/6 satisfy sin(x) = -1/2 in the interval 0 ≤ x ≤ 2π.
Solving the second equation:
sin(x) - 1 = 0
sin(x) = 1
Again, using the unit circle or a reference angle, we find that x = π/2 satisfies sin(x) = 1 in the interval 0 ≤ x ≤ 2π.
Therefore, the solutions to the equation csc^2(x) = 2 + csc(x) in the interval 0 ≤ x ≤ 2π are:
x = 7π/6, 11π/6, and π/2.