How do you solve 2sin^2x+sinx=1?

1 Answer
Nghi N. · Daniel L.
May 5, 2015

y = 2t^2 + t - 1 = 0. Call sin x = t. Solve the quadratic equation.

Since a - b + c = 0, then one real root is (-1) and the other is (-c/a = 1/2).
Solve sin x = -1 -> x = (3pi)/2

Solve sin x = 1/2 -> x = pi/6
This answer would be enough if you were to find the solutions in a given range (for example when x in <0;2pi>) If you are not given such range you must give all solutions remembering that trigonometric functions are periodical.

So it would be

x = (3pi)/2 +2kpi where k in ZZ
and
x = pi/6 +2kpi where k in ZZ

Check:

x = (3pi)/2 -> y = 2 - 1 - 1 = 0 Correct

x = pi/6 -> y = 1/2 + 1/2 - 1 = 0 Correct.

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