Question

Question #c6dcd

Answer 1

`x = 2kpi`
`x = pi/2 + 2kpi`.

Explanation:

Use trig identity:
`sin x + cos x = sqrt2sin (x + pi/4)`
In this case:
`sin x + cos x = sqrt2sin (x + pi/4) = 1`
`sin (x + pi/4) = 1/sqrt2 = sqrt2/2`
Two solutions:
`(x + pi/4) = pi/4 + 2kpi`, and `(x + pi/4) = (3pi)/4 + 2kpi`
a. `x + pi/4 = pi/4 + 2kpi`
`x = 0 + 2kpi`
b. `x + pi/4 = (3pi)/4 + 2kpi`
`x = (3pi)/4 - pi/4 = pi/2 + 2kpi.`

Check.
x = 0 --> sin o = 0 --> cos 0 = 1 --> 0 + 1 = 1 OK
`x = pi/2` --> `sin (pi/2) = 1` --> `cos (pi/2)` = 0 --> 1 + 0 = 1. OK