How do I solve csc^2 x = 5 algebraically over the interval 0 <= x < 2pi?
well, you can rewrite your equation as
sin x = ±1/√5
so, find your reference angle, and use it in each quadrant.
Is there supposed to be a different method since it says "algebraically"?
To solve the equation csc^2(x) = 5 algebraically over the interval 0 <= x < 2pi, you can follow these steps:
1. Start by rewriting the equation as sin^2(x) = 1/5 since csc(x) is the reciprocal of sin(x).
2. Next, take the square root of both sides to solve for sin(x). You will have two possibilities: sin(x) = sqrt(1/5) or sin(x) = -sqrt(1/5).
3. In order to find the solutions between 0 and 2pi, you need to examine the unit circle and find angles with those sine values.
a) First, find the reference angle. You can use the inverse sine function (sin^(-1)) on the absolute value of the sqrt(1/5) or -sqrt(1/5) to get the reference angle.
b) Once you have the reference angle, check the quadrant(s) in which the angle lies (since sine is positive or negative in different quadrants).
c) Then, use the unit circle to find the corresponding angles for the reference angle in the appropriate quadrant(s).
4. Keep in mind that for sin(x) = sqrt(1/5), there will be two solutions (180 degrees apart) since sine is positive in both Quadrant I and Quadrant II.
5. Finally, write down all the solutions in the interval 0 <= x < 2pi that you obtained from steps 3 and 4.
By following these steps, you can solve the equation csc^2(x) = 5 algebraically over the given interval.