Solve for x

Question

a) cos² x + sin x - 3 = 0, XE[0,2PIE]

b)2cos2x-1=0, XE[0, PIE]

Answer 1

(b). 2cos(2x) - 1 = 0

 

         cos(2x) = 1/2

 

         2x = π/3 + 2kπ or 2x = 5π/3 + 2kπ, where k is any integer

 

           x = π/6 + kπ  or 5π/6 + kπ

 

If k = 0, we get x = π/6 or 5π/6

If k = 1, we get x = 7π/6 or 11π/6

 

Other choices for k give values of x that lie outside of [0, 2π]

 

Solution set:  {π/6, 5π/6, 7π/6, 11π/6}

Answer 2

a)

 

Using the trig identity     sin²x + cos²x = 1

 

1 - sin²x + sinx - 3 = 0

 

-sin²x + sinx - 2 = 0

 

 

This is similar to a quadratic equation.

 

Let q = sinx

 

 

-q² + q - 2 = 0

 

 

Using the quadratic formula:

 

q = (-1 ± √(1 - 4(2))) / -2

 

q = (-1 ± √(-7)) / -2

 

 

Since the discriminant is negative, we will not have any real solutions.

 

 

 

 

b)

 

Let G = 2x

 

2cosG - 1 = 0

 

2cosG = 1

 

cosG = 1/2

 

 

G = π/3              and             G = 2π - (π/3)

                                            G = 5π/3

 

             so

 

 

2x = π/3             and          2x = 5π/3

 

 x = π/6              and           x = 5π/6