1 Answer

Bill Mei · Answer 1 · 2022-02-26T15:24:51+0000

Answer:

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

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:

$\displaystyle u=-(1)/(2)\text{ or } u=-1$

And by substitution:

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

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:

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

Where n is an integer.

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

$\displaystyle x=(3\pi)/(2)+2n\pi$

All together, our solutions are:

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

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