How do you solve #cscx+cotx=1# and find all solutions in the interval #[0,2pi)#?
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Given equation
Now we know
Adding (1) & (2) we get
Again Subtracting (2) from (1)
Alternative
Given equation
Again
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To solve the equation csc(x) + cot(x) = 1 and find all solutions in the interval [0, 2π), follow these steps:

Rewrite csc(x) and cot(x) in terms of sine and cosine: csc(x) = 1/sin(x) cot(x) = cos(x)/sin(x)

Substitute these expressions into the equation: 1/sin(x) + cos(x)/sin(x) = 1

Combine the fractions on the left side of the equation: (1 + cos(x))/sin(x) = 1

Cross multiply to eliminate the fraction: 1 + cos(x) = sin(x)

Rearrange the equation: sin(x)  cos(x)  1 = 0

Use trigonometric identities to rewrite sin(x) and cos(x): sin(x) = sqrt(1  cos^2(x)) cos(x) = sqrt(1  sin^2(x))

Substitute these expressions into the equation: sqrt(1  cos^2(x))  cos(x)  1 = 0

Square both sides to eliminate the square roots: 1  cos^2(x)  2cos(x) + cos^2(x)  1 = 0

Simplify the equation: 2cos(x) = 0

Solve for cos(x): cos(x) = 0

Find the solutions for x in the interval [0, 2π) where cos(x) = 0: x = π/2, 3π/2
Therefore, the solutions to the equation csc(x) + cot(x) = 1 in the interval [0, 2π) are x = π/2 and x = 3π/2.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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