PlEASE help me with this!! I need it as soon as possible.......

cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi.
When I slove this, I got
cosx+sinx=(1/ square root of 2).
then I don't know what to do the rest. Please help me.

if you get to cosx+sinx=1/square root of 2, well we are doing identities in my pre-cal class and cosx+sinx always equals 1, so it would be 1/square root of 2.

Sorry, cosx+ sinx is not equal to 1 (from the identites). But, it was like this:

(sinx)^2+ (cosx)^2= 1

Can Anybody answer this question for me???? Please???

PlEASE help me with this!! I need it as soon as possible.......
cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi

square both side:

(cosx+sinx)^x=(1/square root of 2)
then solve: it will work.

To solve the equation cos(x - (π/4)) + sin(x - (π/4)) = 1, you first need to simplify the equation.

1. Start by using the identity for the sine and cosine of the difference of two angles:
cos(A - B) = cosA * cosB + sinA * sinB
sin(A - B) = sinA * cosB - cosA * sinB

2. Apply these identities to the given equation:

cos(x - (π/4)) + sin(x - (π/4))
= cos(x) * cos(π/4) + sin(x) * sin(π/4) + sin(x) * cos(π/4) - cos(x) * sin(π/4)
= (cos(x) + sin(x)) * cos(π/4) + (sin(x) - cos(x)) * sin(π/4)

3. Simplify cos(π/4) and sin(π/4):

cos(π/4) = √2/2
sin(π/4) = √2/2

4. Substitute these values back into the equation:

(cos(x) + sin(x)) * (√2/2) + (sin(x) - cos(x)) * (√2/2) = 1

5. Rearrange the equation by combining like terms:

(√2/2) * cos(x) + (√2/2) * sin(x) - (√2/2) * cos(x) - (√2/2) * sin(x) = 1

Simplifying further, you get:

0 = 1

6. Since the equation simplifies to an invalid statement (0 = 1), there are no solutions to the original equation cos(x - (π/4)) + sin(x - (π/4)) = 1 within the interval [0, 2π].