How do you find all values of x in the interval [0, 2pi] in the equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x - sinx - 1 = 0$ ?

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How do you find all values of x in the interval [0, 2pi] in the equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x - sinx - 1 = 0$ ?

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Answer 1 · 2016-03-27T09:13:38

Solution set in the interval $[0, 2 π]$ $[0,2pi]$ is ${\frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}}$ ${pi/2,(7pi)/6,(11pi)/6}$

Explanation:

To find all values of x in the interval $[0, 2 π]$ $[0, 2pi]$ in the equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x−sinx−1=0$ , let us solve by considering $sin x$ $sinx$ as variable and using quadratic formula.

Hence $sin x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 2 \times (- 1)}}{2 \times 2}$ $sinx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)$ or

$sin x = \frac{1 \pm \sqrt{1 + 8}}{4}$ $sinx=(1+-sqrt(1+8))/4$ i.e.

$sin x = \frac{1 \pm 3}{4}$ $sinx=(1+-3)/4$ i.e. $sin x = 1$ $sinx=1$ or $- \frac{1}{2}$ $-1/2$

In the interval $[0, 2 π]$ $[0,2pi]$ , $sin x = 1$ $sinx=1$ only for $x = \frac{π}{2}$ $x=pi/2$

and $sin x = - \frac{1}{2}$ $sinx=-1/2$ for $x = \frac{7 π}{6}$ $x=(7pi)/6$ and $\frac{11 π}{6}$ $(11pi)/6$

Hence solution set in the interval $[0, 2 π]$ $[0,2pi]$ is ${\frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}}$ ${pi/2,(7pi)/6,(11pi)/6}$

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How do you find all values of x in the interval [0, 2pi] in the equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x - sinx - 1 = 0$ ?

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Explanation:

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How do you find all values of x in the interval [0, 2pi] in the equation $2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x - sinx - 1 = 0$ ?