2^{\sin(x)+\cos(y)}=1, \ \ 16^{\sin^2(x)+\cos^2(y)}=4 (system of equations)

tearstreakdl

tearstreakdl

Answered question

2022-01-16

2sin(x)+cos(y)=1,  16sin2(x)+cos2(y)=4 (system of equations)

Answer & Explanation

GaceCoect5v

GaceCoect5v

Beginner2022-01-17Added 26 answers

Suppose sinx+cosy=0. Then
sin2x+cos2y=(sinx+cosy)22sinxcosy=2sinxcosy
Thus we have left to consider
{sinx+cosy=0sinxcosy=14
raefx88y

raefx88y

Beginner2022-01-18Added 26 answers

From the first equation you get
sinx=cosy
Substituting that in the second equation, you get,
12=sin2x+cos2y=2sin2x
cos2x=12sin2x=12=cosπ3
2x=2nπ±π3
x=nπ±π6   nZ
or you could have directly got that using the fact that
sin2x=sin2π6x=nπ±π6
Similarly, for y, we have
cos2y=cos2π3
y=nπ±π3    nZ

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