Question

Find all real numbers in interval [0,2π)? sin x cos(π/4)+cos x sin (π/4)=1/2

Answer 1

I get `x\in \{{7\pi}/{12},{23\pi}/{12}\}`.

Explanation:

The expression given fits the formula for the sine of a sum:

`\sin(x+y)=\sin(x)\cos(y)\+cos(x)\sin(y)`

Here put in `y={\pi}/{4}` and `\sin(x+y)={1}/{2}`. Then we have the following possibilities:

(1) `x+{\pi}/{4}={\pi}/{6}+2m\pi`

(2) `x+{\pi}/{4}={5\pi}/{6}+2n\pi`

Possibility (1) gives

`x=-{\pi}/{12}+2m\pi`

and to get a number between `0` and `2\pi` we take `m=1`, thus `x={23\pi}/{12}`.

Possibility (2) gives

`x={7\pi}/{12}+2n\pi`

and to get a number between `0` and `2\pi` we take `n=0`, thus `x={7\pi}/{12}`.