Sin²x + 4sinx - 2 = 0, general solution

help? :]

This is what is known as a Special Quadratic equation. These are solved in the same manner as any other quadratic equation


sin(x)^2 + 4sin(x) - 2 = 0
sin(x) = (-4 +/- sqrt(4^2 - 4 * 1 * (-2))) / (2 * 1)
sin(x) = (-4 +/- sqrt(16 + 8)) / 2
sin(x) = (-4 +/- sqrt(24)) / 2
sin(x) = (-4 +/- 2 * sqrt(6)) / 2
sin(x) = -2 +/- sqrt(6)

Now, -2 - sqrt(6) will be far outside the range of the sine function, so it's extraneous

sin(x) = sqrt(6) - 2
x = arcsin(sqrt(6) - 2)
x = 26.710951089979093842467770692683 + 360 * k , 153.28904891002090615753222930732 + 360 * k

In radians:

x = 0.4661940428593033415826129538331 + 2pi * k , 2.6753986107304898968800304294464 + 2pi * k

Where k is an integer

since sin²x + 4sinx - 2 = 0
(sinx+2)²-2-4=0
(sinx+2)²=6

so sinx+2=6^1/2 or sinx+2=-6^1/2
sinx=-2+6^1/2 or sinx=-2-6^1/2

but |sinx|<=1
so sinx=-2+6^1/2