How do you solve ${sin}^{2} x + 2 sin x - 3 = 0$ $sin^2x + 2sinx - 3 = 0$ ?

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Answer 1 · 2015-05-03T03:38:12

Let sin x = t,

$f (t) = t^{2} + 2 t - 3 = 0$ $f(t) = t^2 + 2t - 3 = 0$

Since a + b + c = 0, one real root is t = 1 and the other: $t = \frac{c}{a} = - 3 .$ $t = c/a = -3.$

$S o l v e : t = sin x = 1 = sin (\frac{π}{2}) \to x = \frac{π}{2}$ $Solve: t = sin x = 1 = sin (pi/2) -> x = pi/2$

The solution t = -3 is rejected.

How do you solve sin^2x + 2sinx - 3 = 0sin2x+2sinx−3=0sin^2x + 2sinx - 3 = 0?