How do you solve for x: #sin(x) + 2 sin^2(x) = 1#?
1 Answer
Apr 15, 2015
f(x) = sin x + 2sin^2 x - 1 = 0. Call sin x = t, we get:
f(t) = 2t^2 + t - 1 = 0. This is a quadratic equation, with a - b + c = 0, then one real root is (-1) and the other is (-c/a) = 1/2.
Next, solve the 2 basic trig equation:
1. t = sin x = -1 --> x = 3Pi/2
2. t = sin x = 1/2 --> x = Pi/6 and x = 5Pi/6.
Answers within period (0, 2Pi): Pi/6, 5Pi/6; and 3Pi/2.
Extended answers:
x = Pi/6 + k2Pi
x = 5Pi/6 + k2Pi
x = 3Pi/2 + k2Pi.
Check:
x = Pi/6 --> sin x = 1/2--> f(x) = 1/2 + 2(1/4) - 1 = 0. Correct.
x = 5Pi/6 --> sin x = 1/2 --> f(x) = 1/2 + 2*(1/4) - 1 = 0. Correct
x = 3Pi/2 --> sin x = -1 --> f(x) = -1 + 2(-1)^2 - 1 = 1 + 2 - 1 = 0. Correct
