Question

Find the solution on the interval [0, 2pi)

Answer 1

Recall the angle sum identities:

`\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b`

`\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b`

So

`\cos\left(\frac\pi4-x\right)=\frac1{\sqrt2}\cos x+\frac1{\sqrt2}\sin x`

`\sin\left(\frac\pi4-x\right)=\frac1{\sqrt2}\cos x-\frac1{\sqrt2}\sin x`

Then the left hand side of the equation reduces significantly to give

`\frac2{\sqrt2}\sin x=-1\implies\sin x=-\frac{\sqrt2}2=-\frac1{\sqrt2}`

In general, this occurs for
`x=-\frac\pi4+2n\pi` and
`x=-\frac{3\pi}4+2n\pi` where
`n` is any integer. We get solutions in the interval
`[0,2\pi)` for
`n=1`, for which we get

`x=\frac{5\pi}4,\frac{7\pi}4`