Question

How do you solve `sin(x+pi/4)+sin(x-pi/4)=1` over the interval `(0,2pi)`?

Answer 1

Use the trig identity: `color(blue)(sin (a + b) + sin (a - b) = 2sin a*cos b)`

`f(x) = 2*sin x*cos (pi/4) - 1 = 0` ,

since `color(blue)(cos (pi/4) = (sqrt2)/2`

`f(x) = (sqrt2*sin x) - 1 = 0`

`sin x = 1/sqrt2 = (sqrt2)/2 -->`

`color(red)(x = pi/4 and 3pi/4` (inside interval `0 - 2pi`)

Check:
`x = pi/4 --> x + pi/4 = pi/2 --> sin (x + pi/4) = 1; cos (x + pi/4) = 0 --> f(x) = 1 - 1 = 0`. Correct.
`x = 3pi/4 --> (x + pi/4) = pi --> sin pi = 0; cos pi = -1 --> f(x) = 1 - 1 = 0.` Correct