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Question Number 204518 by Abdullahrussell last updated on 20/Feb/24
Answered by TonyCWX08 last updated on 20/Feb/24
letussolveyy=2firstyy=2y=21yy=(eln(2))1yy=eln(2)yln(2)=ln(2)yeln(2)yUsingLambertWfunction,W(ln(2))=ln(2)yy=ln(2)W(ln(2))y=eW(ln(2))AccordingtoWolframAlpha,eW(ln(2))≈1.55961=ySoyy=21.559611.55961=2tan(x)tan(x)=1.559611.55961tan(x)=1.55961x=tan−1(1.55961)x=1.00064+kπ,k∈Z
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