How do you solve $2 cos^2 x = 3 sin x$ on the interval $[0,2pi]$ ?

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Answer 1 · 2015-07-25T16:08:50

Solve $2cos^2 x = 3sin x$

Ans: $pi/6$ and $(5pi)/6$

$2(1 - sin^2 x) = 3sin x.$ Call sin x = t
$2 - 2t^2 = 3t$ --> $2t^2 + 3t - 2 = 0$
D = d^2 = 9 + 16 = 25 --> d = +- 5
$t = sin x = -3/4 +- 5/4$

$t = 2/4 = 1/2$ and $t = -8/4 = - 2$ (rejected because < -1)

$t = sin x = 1/2$ --> $x = pi/6$ and $x = (5pi)/6$

How do you solve 2 cos^2 x = 3 sin x2 cos^2 x = 3 sin x on the interval [0,2pi][0,2pi]?