How do you solve $6sin^2x + sinx -1 = 0$ ?

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Answer 1 · 2015-05-10T00:15:03

Call sin x = t. Solve quadratic equation:
f(t) = 6t^2 + t - 1 = 0

$D = d^2 = 1 = 24 = 25 -> d = +- 5$

$t1 = -1/12 + 5/12 = 4/ 12 = 1/3$
$t2 = -1/12 - 5/12 = -6/12 = -1/2$

On trig unit circle:

$sin x = t = 1/3 -> x = 19. 47 and x = 199.47$ deg

$sin x = t = -1/2 -> x = (7pi)/6 and x = (11pi)/6$

How do you solve 6sin^2x + sinx -1 = 06sin^2x + sinx -1 = 0?