Solve sin x + cos x = sqrt2sinx+cosx=2 ?

3 Answers

x= pi/4+ 2k pix=π4+2kπ, 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