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

1 Answer
Apr 15, 2018

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}.