Question

How do you solve `\cos x + \sin x = \sqrt { 2}`?

Answer 1

The solution is `S={pi/4+2pin}`, `n in ZZ`

Explanation:

We need

`sin(x+a)=sinxcosa+sinacosa`

Our equation is

`sinx+cosx=sqrt2`

`1/sqrt2sinx+1/sqrt2cosx=1`

Therefore,

`cosa=1/sqrt2`

and `sina=1/sqrt2`

So,

`a=pi/4`

`sin(x+pi/4)=1`

`x+pi/4=pi/2+2pin`

`x=pi/4+2pin`, `nin ZZ`

Answer 2

`x = pi/4 + 2kpi`

Explanation:

Use trig identity:
`sin x + cos x = sqrt2cos (x - pi/4)`
In this case:
`sin x + cos x = sqrt2 = sqrt2cos (x - pi/4)`
`cos (x - pi/4) = 1`
Unit circle gives:
`x - pi/4 = 0` --> `x = pi/4 + 2kpi`
Check by calculator:
`x = pi/4` --> `cos x = sqrt2/2` --> `sin x = sqrt2/2`
`cos x + sin x = sqrt2/2 + sqrt2/2 = sqrt2` (proved)