the answer and process
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√1/ [(|x|-1)cos~1(2x+1).tan3x
so buddy we can solve your question by solving your question in parts
so in
first case :-
√1/ [(|x|-1)cos~1(2x+1).tan3x ≥ 0
in second case :-
|x|-1 > 0
which means x²-1>0
(x-1)(x+1)>0
which means we have to use wavy curve method
as shown in figure i made for u in paint
x∈(-∞,-1) ∪ (1 ,∞)
now in third case :-
The domain of the inverse cosine function is [−1,1]
we can use this
-1≤cosinverse(2x+3) ≤1 but ≠ 0 in any case
so
2x+3≥-1
x≥-2
and
2x+3= 1
x≤ 1
here we find that x belongs to [-2,1} but ≠0
and in tan3x domain is R we can take any number now i will put all the parts of x value and the common part will be our answer dude :-
so dude finally we get the domain of the function
x∈ [-2,-1] ∪[1,∞]
so buddy we can solve your question by solving your question in parts
so in
first case :-
√1/ [(|x|-1)cos~1(2x+1).tan3x ≥ 0
in second case :-
|x|-1 > 0
which means x²-1>0
(x-1)(x+1)>0
which means we have to use wavy curve method
as shown in figure i made for u in paint
x∈(-∞,-1) ∪ (1 ,∞)
now in third case :-
The domain of the inverse cosine function is [−1,1]
we can use this
-1≤cosinverse(2x+3) ≤1 but ≠ 0 in any case
so
2x+3≥-1
x≥-2
and
2x+3= 1
x≤ 1
here we find that x belongs to [-2,1} but ≠0
and in tan3x domain is R we can take any number now i will put all the parts of x value and the common part will be our answer dude :-
so dude finally we get the domain of the function
x∈ [-2,-1] ∪[1,∞]
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