# How do you solve 2sin^2x-5sinx-3 = 0?

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Nghi N Share
May 22, 2018

$x = \frac{7 \pi}{6} + 2 k \pi$
$x = \frac{11 \pi}{6} + 2 k \pi$

#### Explanation:

$f \left(x\right) = 2 {\sin}^{2} x - 5 \sin x - 3 = 0$
Solve this quadratic equation for sin x.
$D = {d}^{2} = {b}^{2} - 4 a c = 25 + 24 = 49$ --> $d = \pm 7$
There are 2 real roots:
$\sin x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{5}{4} \pm \frac{7}{4}$ -->
$\sin x = - \frac{2}{4} = - \frac{1}{2}$, and sin x = 3 (rejected as > 1)
$\sin x = - \frac{1}{2}$
Trig table and unit circle give 2 solutions for x:
$x = - \frac{\pi}{6} + 2 k \pi$, or $x = \frac{11 \pi}{6} + 2 k \pi$ (co-terminal)
and $x = \pi - \left(- \frac{\pi}{6}\right) = \frac{7 \pi}{6} + 2 k \pi$

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