Question

How do you solve `Cos(theta) - sin(theta) = sqrt 2`?

Answer 1

`S={-45^@+360^@n}`

Explanation:

Use the following formulas to transform the equation then solve
`Acosx+Bsinx=C cos(x-D)`
`C=sqrt(x^2+y^2), cosD=A/C, sinD=B/C`

`A=1, B=-1, C=sqrt2`
`cosD=1/sqrt2->D=cos^-1(1/sqrt2)=-45^@`

Note that cosine is positive but sine is negative therefore the triangle is in quadrant four so our angle D will either be `-45^@ or 315^@`. I pick the `-45^@`

`sqrt2 cos(theta+45^@)=sqrt2`

`cos(theta+45^@)=1`

`theta+45^@+=cos^-1 1`

`theta+45^@=0^@+360^@n`

`theta=-45^@+360^@n`

`S={-45^@+360^@n}`