# Solve the equation on the interval [0,2pi] sin{{left({x}+frac{pi}{{4}}right)}}+ sin{{left({x}-frac{pi}{{4}}right)}}={1}

Question
Solve the equation on the interval [0,2pi] $$\sin{{\left({x}+\frac{\pi}{{4}}\right)}}+ \sin{{\left({x}-\frac{\pi}{{4}}\right)}}={1}$$

2020-12-16
Use the sum and difference identities for sine:
$$\sin{{\left({A}\pm{B}\right)}}= \sin{{A}} \cos{{B}}\pm \cos{{A}} \sin{{B}}$$
$${\left( \sin{{x}}\frac \cos{\pi}{{4}}+ \cos{{x}}\frac \sin{\pi}{{4}}\right)}+{\left( \sin{{x}}\frac \cos{\pi}{{4}}+ \cos{{x}}\frac \sin{\pi}{{4}}\right)}=1$$
$${2}\sin{{x}}\frac \cos{\pi}{{4}}=1$$
$${2}\sin{{x}}\cdot\frac{\sqrt{{2}}}{{2}}=1$$
$$\sqrt{{2}} \sin{{x}}=1$$
$$\sin{{x}}=\frac{1}{\sqrt{{2}}}=\frac{\sqrt{{2}}}{{2}}$$
Sine is positive on QI and QII.
The reference angle is $$\frac{\pi}{{4}} ,\ \text{since} \ \frac \sin{\pi}{{4}}=\frac{\sqrt{{2}}}{{2}}$$
The QI solution is:
$${x}=\frac{\pi}{{4}}$$
The QII solution is:
{x}=\pi-\frac{\pi}{{4}}=\frac{{{3}\pi}}{{4}}\)

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