Question

Solve `sin x + cos x = sqrt2` ?

Answer 1

`x= pi/4+ 2k pi`, with `k in ZZ`

Explanation:

Oke, I can't come up with anything simpler than this...

`cosx + sinx = sqrt2`
`sin(x+pi/2) + sinx = sqrt2`

Now we know that: `sin(a+b) + sin(a-b) = 2 sina cosb`. To use this equation, we say for example:

`a+b = x+pi/2`
`a-b = x`

Solving gives:

`a = x + pi/4`
`b = pi/4`

So now we get:

`sin(x+pi/2) + sinx = sin(a+b) + sin(a-b) = 2 sina cosb = 2sin(x+pi/4) cos(pi/4) = 2sin(x+pi/4) sqrt2 /2 = sqrt2 sin(x+pi/4)`

Now the equation gets much simpler:

`sinx + sin(x+pi/2) = sqrt2`
`sqrt2 sin(x+pi/4) = sqrt2`
`sin(x+pi/4) = 1`
`x+pi/4 = pi/2 + 2k pi`
`x= pi/4+ 2k pi`

Where `k in ZZ`

Answer 2

The answer is `pi/4+2kpi`.

Explanation:

![https://useruploads.socratic.org/RMhT5Io8TTuxYQlIKWRt_1501808602898-1613214639.jpg)

Answer 3

`x = pi/4+2kpi, k=0,pm1,pm2,cdots`

Explanation:

Calling `y= sinx` we have

`y+sqrt(1-y^2) = sqrt2` or `sqrt(1-y^2)=sqrt2-y`

now squaring both sides

`1-y^2=2-2sqrt2y+y^2` or `2y^2-2sqrt2y+1=0` or

`y^2-sqrt2y+1/2=0` or

`(y-1/sqrt 2)^2=0 rArr y=1/sqrt2=sinx rArr x = pi/4+2kpi, k=0,pm1,pm2,cdots`