Question

Question #92a25

Answer 1

`x=0,pi/2,2pi`

Explanation:

`1-sinx=cosx`

`=>cosx+sinx=1`

`=>1/sqrt2cosx+1/sqrt2sinx=1/sqrt2`

`=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi/4)`

`=>cos(x-pi/4)=cos(pi/4)=cos(-pi/4)=cos(2pi-pi/4)`

So `x-pi/4=pi/4`

`=>x=pi/2`

when

`x-pi/4=-pi/4`

`x=0`

Again

`x-pi/4=2pi-pi/4`

`=>x=2pi`

Answer 2

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

Explanation:

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