Sin^2(x)−3sin(x) = −1...how do i find all solutions?

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Sin^2(x)−3sin(x) = −1...how do i find all solutions?

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Answer 1 · 2018-05-13T13:01:33

$x = 2 n π + {(- 1)}^{n} {sin}^{- 1} (\frac{3 - \sqrt{5}}{2})$ $x=2npi+(-1)^nsin^-1((3-sqrt5)/2)$

Explanation:

${sin}^{2} x - 3 sin x = - 1$ $sin^2x-3sinx=-1$
let $sin x = p$ $sinx=p$
then equation becomes
$p^{2} - 3 p = - 1$ $p^2-3p=-1$ now solve for $p$ $p$
as,
$p^{2} - 3 p = - 1$ $p^2-3p=-1$
$\Rightarrow p^{2} - 3 p + 1 = 0$ $=>p^2-3p+1=0$
from quadratic formula
$p = \frac{3 \pm \sqrt{{(- 3)}^{2} - 4 (1) (1)}}{2 (1)}$ $p=(3+-sqrt((-3)^2-4(1)(1)))/(2(1))$
$\Rightarrow p = \frac{3 + \sqrt{5}}{2}$ $=>p=(3+sqrt5)/2$ or $p = \frac{3 - \sqrt{5}}{2}$ $p=(3-sqrt5)/2$
$=>sinx=(3-sqrt5)/2,cancel((3+sqrt5)/2)$ ,second value not possible since $sinx$ cannot exceed 1
$(3-sqrt5)/2$ can be written as $sin(sin^-1((3-sqrt5)/2))$
$=:.sinx=sin(sin^-1((3-sqrt5)/2))$
$=>x=2npi+(-1)^nsin^-1((3-sqrt5)/2)$
this is general solution
put $n=0,1$ for principal solution

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Sin^2(x)−3sin(x) = −1...how do i find all solutions?

1 Answer

Explanation:

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