Solve \(\displaystyle{{\tan}^{{2}}{\left({x}\right)}}-{\left({1}+\sqrt{{3}}\right)}{\tan{{\left({x}\right)}}}+\sqrt{{3}}{ < }{0}\) But I get

ropowiec2gkc

ropowiec2gkc

Answered question

2022-03-21

Solve tan2(x)(1+3)tan(x)+3<0
But I get D=423 and I don't think that's the right way to go about this.

Answer & Explanation

cineworld93uowb

cineworld93uowb

Beginner2022-03-22Added 16 answers

Explanation:
How about changing the question?
(tanx1)(tanx3)<0
So the zeroes of that function are π4 and π3.
If you look at the values in brackets, both are negative for x<π4, meaning the final value is positive. For x>π3 both values in brackets are positive, meaning the final value is also positive.
Let's analyse that interval x(dπ4,dπ3). In this interval, (tanx1) is positive but (tanx3) is negative, meaning the final result is negative and hence <0.
So here's your final answer:
dπ4+2πn<x<dπ3+2πnnN

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