Trigonometric equation: \(\displaystyle{{\ln{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}}^{{{1}+{\sin{{2}}}{x}}}=}{2}\)

Aldo Fernandez

Aldo Fernandez

Answered question

2022-04-10

Trigonometric equation: ln(sinx+cosx)1+sin2x=2

Answer & Explanation

Ouhamiptkg

Ouhamiptkg

Beginner2022-04-11Added 18 answers

HINT
ln(sinx+cosx)1+sin2x=2ln(sinx+cosx)1+sin2x=lne2(sinx+cosx)1+sin2x=e2
but
(sinx+cosx)1+sin2x=(sinx+cosx)(sinx+cosx)222=2
Note
(sinx+cosx)2=1+sin2x2sinx+cosx2

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