Question #aa66a

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Answer 1 · 2017-11-17T06:24:11

Write this as follows

$sinx-cosx=1$

Multiply with $sqrt2/2$ both sides

$sqrt2/2*(sinx-cosx)=sqrt2/2$

$sin(x-pi/4)=sin(pi/4)$

Hence the general solutions are

$x-pi/4=2*κ*pi+pi/4$ and $x-pi/4=2*k*pi+3pi/4$

or

$χ=2*k*pi+pi/2$ and $x=2*k*pi+pi$

Because x belongs in $(0,2*pi]$ acceptable solutions are

$x=pi$ and $x=pi/2$