How do you solve sinx+cosx=1?

3 Answers
Jun 24, 2016

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]}

Feb 21, 2017

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...

Feb 24, 2017

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