How do I find all solutions in the interval [0, 2Pi):
cotx=cscx-1
the answer choices are:
a.) x=Pi/2
b.) x=Pi/2 and 3Pi/2
pi/2 surely works, (try subbing it in)
remember that 3pi/2 is 270º
tan 270º is undefined, so cot 270º = 0
LS = cot 3pi/2 = 0
RS = 1/sin 270 - 1 = 1/-1 - 1 = -2
so 3pi/2 does not work.
To find all solutions to the equation cot(x) = csc(x) - 1 in the interval [0, 2π), we will need to apply some trigonometric identities and algebraic manipulation.
1. Start by converting cot(x) and csc(x) to their equivalent forms using sine and cosine.
cot(x) = csc(x) - 1
cos(x)/sin(x) = 1/sin(x) - 1
2. Multiply both sides of the equation by sin(x) to eliminate the denominators.
cos(x) = 1 - sin(x)
3. Square both sides of the equation to get rid of the square root.
cos^2(x) = (1 - sin(x))^2
cos^2(x) = 1 - 2sin(x) + sin^2(x)
4. Recognize that cos^2(x) = 1 - sin^2(x) due to the Pythagorean identity.
1 - sin^2(x) = 1 - 2sin(x) + sin^2(x)
5. Simplify the equation.
2sin^2(x) - 2sin(x) = 0
6. Factor out sin(x).
2sin(x)(sin(x) - 1) = 0
7. Set each factor equal to zero and solve for x.
a) sin(x) = 0
To find the solutions to this equation, we look for all angles x where sin(x) = 0 in the interval [0, 2π).
The solutions for sin(x) = 0 are x = 0, π, and 2π. However, we need to check if the solutions lie within the given interval [0, 2π). Therefore, x = 0 and x = 2π are valid solutions.
b) sin(x) - 1 = 0
To find the solutions to this equation, we solve for x when sin(x) = 1.
From our knowledge of trigonometry, we know that sin(x) = 1 when x = π/2.
Thus, the complete solution set in the interval [0, 2π) is x = 0, π, 2π, and π/2.
Comparing the solution set with the answer choices, we can see that the correct answer is b.) x = π/2 and 3π/2.