How do you solve ${sin}^{2} (x) + sin x - 2 = 0$ $sin^2(x) + sinx - 2 = 0$ ?

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How do you solve ${sin}^{2} (x) + sin x - 2 = 0$ $sin^2(x) + sinx - 2 = 0$ ?

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Answer 1 · 2015-04-15T23:21:37

$f (x) = {sin}^{2} x + sin x - 2 = 0 .$ $f(x) = sin^2 x + sin x - 2 = 0.$ Call $sin x = t$ $sin x = t$ , we get:
$f (t) = t^{2} + t - 2 = 0$ $f(t) = t^2 + t - 2 = 0$ . This is a quadratic equation with $a + b + c = 0$ $a + b + c = 0$ . One real root is ( $1$ $1$ ) and the other is $(\frac{c}{a}) = - 2$ $(c/a) = -2$ (rejected since $> 1$ $> 1$ ).
Next, solve $t = sin x = 1 - \to x = \frac{π}{2}$ $t = sin x = 1 --> x = pi/2$
Answer within period ( $0, 2 π$ $0, 2pi$ ): $x = \frac{π}{2}$ $x = pi/2$
Extended answers: $x = \frac{π}{2} + k \cdot 2 π .$ $x = pi/2 + k*2pi.$
Check:
$x = \frac{π}{2} - \to sin x + 1 - \to f (x) = 1 + 1 - 2 = 0$ $x = pi/2 --> sin x + 1--> f(x) = 1 + 1 - 2 = 0$ . Correct.

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How do you solve ${sin}^{2} (x) + sin x - 2 = 0$ $sin^2(x) + sinx - 2 = 0$ ?

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How do you solve ${sin}^{2} (x) + sin x - 2 = 0$ $sin^2(x) + sinx - 2 = 0$ ?