Question

How do you solve `sinx+cosx=1`?

Answer 1

`x = 2npi and x = (4n+1)pi/2, n = 0, +-1, +-2, +-3. ...`

Explanation:

The given equation is equivalent to

`1/sqrt 2 sin x+ 1/sqrt 2 cosx=1/sqrt 2`.

This can be written as

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

The general solution of this equation ls

`x-pi/4=2npi+-pi/4, n = 0, +-1, +-2, ...`,

So, `x = 2npi and x = (4n+1)pi/2, n = 0, +-1, +-2, +-3. ...`

Having noted that there were 40K viewers for the answers by me,

Hero and Nghi, I think I could invoke more interest by including the

solutions for `cos x - sin x = 1`, and for that matter,

`sec x +- tan x =1`, that become

`cos x - sin x =1` and `cos x + sin x = 1`, upon multiplication by

cos x, when `x ne` an odd multiple of `pi/2`.

For cos x - sin x = 1,

the general solution is

`x = 2npi and x = (4n -1)pi/2, n = 0, +-1, +-2, +-3. ...`

Note the change in the multiple from `( 4n + 1 ) to ( 4n - 1 )`.

For `sec x +- tan x = 1`, it is same sans `(4n +- 1)pi/2`. It is just

`x = 2npi`

See x-intercepts as graphical solutions.

Graph on uniform scale for solutions

`x = ... -2pi, -(3pi)/2, 0, pi/2, 2pi ..` of cos x + sin x = 1:
graph{y-cos x - sin x +1 = 0[-7 7 -3 4]}
Graph on uniform scale for solutions

`x = -2pi, -pi/2, 0, (3pi)/2, 2pi,..` of `cos x - sin x = 1`:
graph{y-cos x + sin x +1 = 0[-7 7 -3 4]}

See combined graph for solutions `0, +-2pi, +-4pi,...` of

`sec x +- tan x = 1`:

graph{(y- sec x - tan x +1)(y- sec x+ tan x +1)=0[-13 13 -6.5 6.5]}

Answer 2

`x=pi/2+2pin or x=2pin` where `n=0,+-1,+-2,+-3...`

Explanation:

If `cosx+sinx=1` then squaring both sides gives us:

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

Using the identities:
`cos^2x+sin^2x=1` and `sin2x=2sinxcosx`

The equation can be simplified to: `1+sin2x=1`

Therefore `sin2x=0`

The values of `theta` at which `sintheta=0` are `theta=npi` where `n` is an integer.

But here `theta=2x` so `x=npi/2` for `n=0,+-1,+-2,+-3...`

However, since we squared the equation, we need to check all of these answers work in the original equation:

`cos0+sin0=1+0=1`

`cos(pi/2)+sin(pi/2)=0+1=1`

`cospi+sinpi=-1+0=-1` This solution is not valid.

`cos(3pi/2)+sin(3pi/2)=0-1=-1` This solution is not valid either.

When we get to `2pi` the graphs repeat, so `2pi,5pi/2,4pi,9pi/2...` are all valid, but the other solutions aren't.

This can be written as:
`x=pi/2+2pin or x=2pin` where `n=0,+-1,+-2,+-3...`

Answer 3

`x = (2k + 1)(pi/2)`
`x = 2kpi`

Explanation:

Use trig identity:
`sin x + cos x = sqrt2cos (x - pi/4)`
We get:
`sqrt2cos (x - pi/4) = 1` --> `cos (x - pi/4) = 1/sqrt2 = sqrt2/2`
Trig table and unit circle give -->
`(x - pi/4) = +- pi/4 + 2kpi`--> 2 solutions -->
`a. x - pi/4 = pi/4 + 2kpi` -->
`x = pi/4 + pi/4 = pi/2 + 2kpi`, and
`b. x - pi/4 = - pi/4 + 2kpi`
`x = pi/4 - pi/4 = 0 + 2kpi = 2kpi`