Question

Solve the equation `sin(x+pi/4)−cos(x)=0`?

Answer 1

`x = pi/8+ npi; n in ZZ`

Explanation:

Given:

`sin(x+pi/4)-cos(x)=0`

Use the identity `sin(A+B) = sin(A)cos(B)+cos(A)sin(B)` where `A = x and B = pi/4`:

`sin(x)cos(pi/4)+cos(x)sin(pi/4)-cos(x) = 0`

Use the fact that `cos(pi/4)=sin(pi/4) = sqrt2/2`:

`sin(x)sqrt2/2+cos(x)sqrt2/2-cos(x) = 0`

Multiply both sides by `sqrt2`:

`sin(x)(sqrt2sqrt2)/2+cos(x)(sqrt2sqrt2)/2-sqrt2cos(x) = 0`

Simplify by observing that `(sqrt2sqrt2)/2 = 1`:

`sin(x)+cos(x)-sqrt2cos(x) = 0`

Divide both sides by `cos(x)`:

`sin(x)/cos(x)+ 1-sqrt2 = 0`

Add `sqrt2-1` to both sides:

`sin(x)/cos(x)=sqrt2 -1`

Use the identity `sin(x)/cos(x) = tan(x)`:

`tan(x)=sqrt2 -1=tan(pi/8)`

`x = pi/8`

The inverse tangent repeats at every integer multiple of `pi`:

`x = pi/8+ npi; n in ZZ`

Answer 2

`x=pi/8+kpi, k in ZZ` , with a simpler resolution

Explanation:

`sin(x+pi/4)−cos(x)=0`

By the properties of sines and cosines

`sin(x+pi/4)=cos(pi/2-x-pi/4)=cos(pi/4-x)`

So the equation turns into:

`cos(pi/4-x)=cos(x)`

`pi/4-x=+-x+2kpi`

`cancel(pi/4=0 +2kpi) or pi/4=2x+2kpi`

`2x=pi/4+2kpi`

`x=pi/8+kpi, k in ZZ`