How do you solve $2sin^2x=sinx+1$ for $0<=x<=2pi$ ?

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Answer 1 · 2017-01-09T09:55:00

$x = pi/2, 7/6pi and 11/6pi$

$2sin^2x = sin x+1$
$2sin^x - sin x - 1 =0$
$(2 sinx +1)(sinx-1) = 0$

$a. 2sinx + 1 =0, sinx = -1/2$
$x = sin^-1(-1/2)$
$x = 7/6(pi),11/6(pi)$

$b. sin-1 =0, sinx = 1$
$x = sin^-1(1)$
$x = pi/2$

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