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find all solutions:
cotx-1=cscx; in the interval [0,2pi)
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Your equation is
- 1 = in the interval [0, 2pi).
Notice that sin(x) is in the denominator: it means that the domain of this equation
is the part of the interval [0,2pi), where sin(x) =/= 0.
In other words, the domain is the union of intervals (0,pi) U (pi,2pi).
The points x= 0 and x= pi are gouged out.
OK. Multiply both sides by sin(x). We can do it in the domain, since sin(x) =/= 0 there.
You will get
cos(x) - sin(x) = 1.
To solve, square both sides
cos^2(x) - 2cos(x)*sin(x) + sin^2(x) = 1.
Replace cos^2(x) + sin^2(x) by 1 and cancel 1 in both sides. You will get
-2cos(x)*sin(x) = 0.
It means that either cos(x) = 0 or sin(x) = 0.
But sin(x) = 0 is excluded in the domain; so, the only possibility to consider is cos(x) = 0.
It gives two solutions for x: x = pi/2 and x= 3pi/2.
Check shows that x= pi/2 is an extraneous solution.
ANSWER. In the given domain, the original equation has a unique solution x= 3pi/2.
Solved.