Solve Equation using Quadratic Formula

Question

Solve the equation using Quadratic Formula and a Calculator. Find the solutions in the interval [o,2pi). Round answers to 4 decimal places.

sin^2 x - sin x -1 = 0

Answer 1

Solve the equation using Quadratic Formula and a Calculator. Find the solutions in the interval [0,2π). Round answers to 4 decimal places. sin²x - sinx -1 = 0

So in Trig quadratic function, you can replace the trig function with an x. You then solve the quadratic, get your x solutions, then bring back the trig function and solve for the angle.

1. In the above example, replace the sinx with just x. (x² - x -1 = 0) a = 1, b = -1, c = -1
2. -b ±√((b²-4ac)/ / (2(a))
3. sin x =( - -1 ± √((-1)² - (4)(1)(-1) )/ (2(1))
4. sin x = 1/2 ± √5/2
5. x = sin^-1(1/2 ± √5/2); x = sin^-1 1.618 not a possible result; x = sin^-1 -.618 = -38.1727°
6. x = -38.1727°; or x = 321.8273°

Does this make sense to you on how you use the quadratic formula in trig functions set up in in standard quadratic form? I hope!

Answer 2

I always find it easier to work with these sorts of equations by making a substitution for sin x which is what I will do here Let sinx = a so equation is a^2-a-1=0. using the quadratic formula I get

the solutions to be [ 1 +/-sqroot(1+4)]/2 = (1+sqroor5)/2 and (1-sqroot5)/2=2.2360 and -.6180 so

sina= -.6180 a= -38.1726 or to express this as a positive angle this would be 321.8274 and

180 +38.1726= 218.1726

The other solution is extraneous since there is no angle with a sin of 2.2360 ( sin varies from -1 to 1 )

Answer 3

let z = sinx

z^2 - z - 1 = 0

z^2 - z + 1/4 = 5/4

z- 1/2 = (sqr5)/2

z= sinx = about -0.6

x = 36.9 degrees = pi/5 radians