Solve #sin x + cos x = sqrt2# ?

3 Answers

# x= pi/4+ 2k pi#, with #k in ZZ #

Explanation:

Oke, I can't come up with anything simpler than this...

#cosx + sinx = sqrt2#
#sin(x+pi/2) + sinx = sqrt2#

Now we know that: #sin(a+b) + sin(a-b) = 2 sina cosb#. To use this equation, we say for example:

#a+b = x+pi/2#
#a-b = x#

Solving gives:

#a = x + pi/4#
#b = pi/4#

So now we get:

#sin(x+pi/2) + sinx = sin(a+b) + sin(a-b) = 2 sina cosb = 2sin(x+pi/4) cos(pi/4) = 2sin(x+pi/4) sqrt2 /2 = sqrt2 sin(x+pi/4)#

Now the equation gets much simpler:

#sinx + sin(x+pi/2) = sqrt2#
#sqrt2 sin(x+pi/4) = sqrt2#
#sin(x+pi/4) = 1#
#x+pi/4 = pi/2 + 2k pi#
# x= pi/4+ 2k pi #

Where #k in ZZ #

The answer is #pi/4+2kpi#.

Explanation:

![https://useruploads.socratic.org/RMhT5Io8TTuxYQlIKWRt_1501808602898-1613214639.jpg)

Aug 4, 2017

# x = pi/4+2kpi, k=0,pm1,pm2,cdots#

Explanation:

Calling #y= sinx# we have

#y+sqrt(1-y^2) = sqrt2# or #sqrt(1-y^2)=sqrt2-y#

now squaring both sides

#1-y^2=2-2sqrt2y+y^2# or #2y^2-2sqrt2y+1=0# or

#y^2-sqrt2y+1/2=0# or

#(y-1/sqrt 2)^2=0 rArr y=1/sqrt2=sinx rArr x = pi/4+2kpi, k=0,pm1,pm2,cdots#