Giải phương trình: $\sin \left(2x+\frac{3\text{π}}{4}\right)+\cos x\text{\hspace{0.17em}= 0}$

Ta có $\sin \left(2x+\frac{3\text{π}}{4}\right)+\cos x\text{\hspace{0.17em}= 0}\iff \cos x=\sin \left(-2x-\frac{3\text{π}}{4}\right)$

$\iff \cos x=\sin \left(\frac{\text{π}}{2}-\left(2x+\frac{\text{5π}}{4}\right)\right)\iff \cos x=\cos \left(2x+\frac{\text{5π}}{4}\right)$

$\iff \left[\begin{array}{l}x=2x+\frac{\text{5π}}{4}+k2\text{π}\\ x=-2x-\frac{\text{5π}}{4}+k2\text{π}\end{array}\right.\iff \left[\begin{array}{l}x=-\frac{\text{5π}}{4}-k2\text{π}\\ x=-\frac{\text{5π}}{12}+\frac{2}{3}k\text{π}\end{array}\right.(k\in {\rm Z})$

Vậy phương trình đã cho có 2 họ nghiệm là: $x=-\frac{\text{5π}}{4}-k2\text{π}$và$x=-\frac{\text{5π}}{12}+\frac{2}{3}k\text{π.}$