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

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

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

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