Answer:
[tex]\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}[/tex]
Step-by-step explanation:
We are given the equation:
[tex]4\sin^2(x)+6\sin(x)+2=0[/tex]
First, we can divide everything by 2:
[tex]2\sin^2(x)+3\sin(x)+1=0[/tex]
Notice that we have an equation in quadratic form. Namely, if we make a substitution where u = sin(x), we acquire:
[tex]2u^2+3u+1=0[/tex]
Solve for u. Factor:
[tex](2u+1)(u+1)=0[/tex]
Zero Product Property:
[tex]2u+1=0\text{ or } u+1=0[/tex]
Solving for both cases:
[tex]\displaystyle u=-\frac{1}{2}\text{ or } u=-1[/tex]
And by substitution:
[tex]\displaystyle \sin(x)=-\frac{1}{2}\text{ or } \sin(x)=-1[/tex]
For the first case, recall that sin(x) is -1/2 for every 7π/6 and every 11π/6. Hence, for the first case, our solutions are:
[tex]\displaystyle x=\frac{7\pi}{6}+2n\pi \text{ and } x=\frac{11\pi}{6}+2n\pi, n\in\mathbb{Z}[/tex]
Where n is an integer.
For the second case, sin(x) is -1 for every 3π/2. Thus:
[tex]\displaystyle x=\frac{3\pi}{2}+2n\pi[/tex]
All together, our solutions are:
[tex]\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}[/tex]
