Find the inverse function of tan2(x)−23tan⁡(x)The question is pretty clear, we must find the inverse function off:x→tan2(x)−23(tan⁡(x))This function is a bijection from (−π2+kπ,π3+kπ) to (−3,+∞) since it is strictly decreasing and continuous.I've tried factorizing with tangent and playing with the equation so I can add Arctangent to eliminate tangent like with usual functions but I do not seem to be capable succeeding using this method.P.S: This function is also a bijection from (−π3+kπ,π2+kπ) to (−3,+∞)

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Find the inverse function of tan2(x)−23tan⁡(x)The question is pretty clear, we must find the inverse function off:x→tan2(x)−23(tan⁡(x))This function is a bijection from (−π2+kπ,π3+kπ) to (−3,+∞) since it is strictly decreasing and continuous.I&#039;ve tried factorizing with tangent and playing with the equation so I can add Arctangent to eliminate tangent like with usual functions but I do not seem to be capable succeeding using this method.P.S: This function is also a bijection from (−π3+kπ,π2+kπ) to (−3,+∞)

vadomosigmx6 · Accepted Answer

Consider[−1,3]→[1+23,−3]x↦x2−23xThen f=p⋅tan⁡ and therefore f−1=tan−1⋅p−1=arctan⋅p−1. Finally, p−1(x)=3−x+3 and thereforef−1(x)=arctan⁡(3−x+3)

Find the inverse function of \(\displaystyle{{\tan}^{{2}}{\left({x}\right)}}-{2}\sqrt{{3}}{\tan{{\left({x}\right)}}}\) The question

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