How do you solve (cscx-cotx)^4(cscx+cotx)^4=1?

Answer 1

Zoe Austin

Infinite solutions

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#(cscx-cotx)^4(cscx+cotx)^4=1#

#((cscx-cotx)(cscx+cotx))^4=1#

#(csc^2x-cot^2x)^4=1#

#csc^2x-cot^2x=+-1#

#csc^2x-cot^2x=1. :. csc^2x=1+cot^2x#, Infinite solutions

#csc^2x-cot^2x=-1, :. 1+cot^2x-cot^2x=-1#, no solution

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Answer 2

Oliver Atkinson

To solve the equation $(\csc x - \cot x)^4(\csc x + \cot x)^4 = 1$ , we can use the fact that if two numbers multiply to give 1, then each of those numbers must be either 1 or -1. So, we can set each factor equal to 1 or -1 and solve for $x$ .

For $(\csc x - \cot x)^4 = 1$ : This implies $\csc x - \cot x = \pm 1$ . Solve each case separately.
For $(\csc x + \cot x)^4 = 1$ : This implies $\csc x + \cot x = \pm 1$ . Solve each case separately.

After solving these cases, you will have multiple solutions for $x$ . These solutions will depend on the range of $x$ you are interested in. Remember that trigonometric functions have periodic behavior, so solutions may repeat after a certain interval.

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Answer from HIX Tutor

When evaluating a one-sided 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.