# Solve the equation for exact solutions over the interval [0,360) for 9sin^2x-6sinx=1 ?

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Feb 9, 2018

#### Answer:

color(blue)(x= 53.58^@ , 126.42^@ , 187.94^@, 352.06^@

#### Explanation:

$9 {\sin}^{2} x - 6 \sin x = 1$

$9 {\sin}^{2} x - 6 \sin x - 1 = 0$

Let $\textcolor{w h i t e}{88} u = \sin x$

Then:

$9 {u}^{2} - 6 u - 1 = 0$

Using Quadratic Formula:

$u = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - \left(4 \cdot 9 \cdot - 1\right)}}{2 \cdot 9}$

$u = \frac{6 \pm \sqrt{36 + 36}}{18} = \frac{6 \pm 6 \sqrt{2}}{18} = \frac{1 \pm \sqrt{2}}{3}$

$u = \sin x = \frac{1 + \sqrt{2}}{3} \implies x = \arcsin \left(\frac{1 + \sqrt{2}}{3}\right) = {53.58}^{\circ} , {126.42}^{\circ}$

And:

$x = \arcsin \left(\frac{1 - \sqrt{2}}{3}\right) = {187.94}^{\circ} , {352.06}^{\circ}$

Solutions:

color(blue)(x= 53.58^@ , 126.42^@ , 187.94^@, 352.06^@

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