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

3 Answers
Jun 24, 2016

Answer:

#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

Answer:

#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

Answer:

#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#