Question

I need to solve the equation please help me

Answer 1

Answer:

$\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}$

Step-by-step explanation:

We are given the equation:

$4\sin^2(x)+6\sin(x)+2=0$

First, we can divide everything by 2:

$2\sin^2(x)+3\sin(x)+1=0$

Notice that we have an equation in quadratic form. Namely, if we make a substitution where u = sin(x), we acquire:

$2u^2+3u+1=0$

Solve for u. Factor:

$(2u+1)(u+1)=0$

Zero Product Property:

$2u+1=0\text{ or } u+1=0$

Solving for both cases:

$\displaystyle u=-\frac{1}{2}\text{ or } u=-1$

And by substitution:

$\displaystyle \sin(x)=-\frac{1}{2}\text{ or } \sin(x)=-1$

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:

$\displaystyle x=\frac{7\pi}{6}+2n\pi \text{ and } x=\frac{11\pi}{6}+2n\pi, n\in\mathbb{Z}$

Where n is an integer.

For the second case, sin(x) is -1 for every 3π/2. Thus:

$\displaystyle x=\frac{3\pi}{2}+2n\pi$

All together, our solutions are:

$\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}$