Question

solve the equation: sin x + cos x= =sqrt2 on the interval 0< x <2pi

Answer 1

I think the first step is to square both sides, so \[\sin x + \cos x = -\sqrt{2}\]
becomes
\[\sin ^{2}x + \cos ^{2}x = -2\]

Answer 2

I have two identities that I think apply:

\[\sin ^{2}x + \cos ^{2} =1\]
and
\[\sin ^{2}x = (1-2cosx)/2\]

Answer 3

the equation likes sin x + cos x = sqrt(2) or sin x + cos x = -sqrt(2) ?

Answer 4

The second one, sorry. Had a typo in my original post.

Answer 5

i'm afraid sin^2x + cos^x is not the square of sin x + cos x so that method won't work

one way to do this is by the complementary angle method

Answer 6

- give me a little time to remind myself of this...

Answer 7

rather its called the auxiliary angle method but i can't find my notes on it

Answer 8

do this, square both sides of the equation, you have :

sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 2
and since sin^2(x) + cos^2(x) = 1 , 2 sin(x)cos(x) = sin(2x)

you will have sin(2x) + 1 = 2...so, sin(2x) = 1...can you solve 2x from here ? and then x ?

Answer 9

i can't use the math notation, so it's a little bit hard to read :), if you have further question, please ask :)