How do you solve #2sin^2 x-sinx-1=0#?

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How do you solve #2sin^2 x-sinx-1=0#?

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Answer 1 · 2015-07-20T04:11:39

Solve #2sin^2 x - sin x - 1 = 0#

Answer: #pi/2; (7pi)/6 and (11pi)/6#

Explanation:

Call sin x = t.
#2t^2 - t - 1 = 0.#
Since a + b + c = 0. use shortcut. 2 real roots: t = 1 and #t = c/a = -1/2#
sin x = t = 1 --> #x = pi/2#
sin x = t = -1/2 --> #x = (7pi)/6# and #x = (11pi)/6#

Within interval #(0, 2pi)#: 3 answers: #pi/2; (7pi)/6 , and (11pi)/6.#

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How do you solve #2sin^2 x-sinx-1=0#?

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