Question

How do you solve `sin x + cos x = 1`?

Answer 1

Refer to explanation

Explanation:

Squaring both sides of the equation yields to

`sinx+cosx=1=>(sinx+cosx)^2=1^2=>sin^2x+cos^2x+2cosxsinx=sin^2x+cos^2x=>sinx*cosx=0=>sinx=0 or cosx=0`

The solutions to `sinx=0 or cosx=0` are `0,90,270,360` but `270` does not satisfy the original equation.
So the solutions are `0^o,90^o,360^o`

The following identities were used

`sin^2x+cos^2x=1`

`(a+b)^2=a^2+b^2+2ab`

Answer 2

Solve sin x + cos x = 1

Ans: x = 0 and `x = pi/2` and `x = 2pi`

Explanation:

Use the trig identity: `sin x + cos x = sqrt2.sin (x + pi/4)`
We get:
`sin (x + pi/4) = sqrt2/2` --> `(x + pi/4) = sin pi/4`
and `(x + pi/4) = sin ((3pi)/4)`

a. `x + pi/4 = pi/4` --> `x = 0` and `x = 2pi`
b. `x + pi/4 = ((3pi)/4)` --> `x = ((3pi)/4) - pi/4 = pi/2`

Check.
x = 0 --> sin x = 0; cos x = 1. Then 0 + 1 = 1. OK
`x = pi/2` --> sin x = 1; cos x = 0. Then 1 + 0 = 1 . OK