Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) 4 cos^2 x + 2 cos x − 2 = 0

Answer 1

Let's throw the cosine function cos(x) into a variable - we'll say y = cos(x). Our expression now becomes

4y² + 2y - 2 = 0

Factoring out the greatest common factor, 2, of the three terms yields the expression

2(2y² + y - 1) = 0

To completely simplify, we can factor the expression 2y² + y - 1 by grouping:

2y² + y - 1

= 2y² + 2y - y - 1

= 2y(y+1) - (y + 1)

= (2y - 1)(y + 1)

Our equation with the left hand side full factored now looks like

2(2y-1)(y+1) = 0

And we can solve for the two cases 2y - 1 = 0 and y + 1 = 0:

2y - 1 = 0 → 2y = 1 → y = 1/2

y + 1 = 0 → y = -1

Finally, we can substitute cos(x) for y, giving us the solutions cos(x) = 1/2 and cos(x) = -1. On the interval [0, 2π), cos(x) = 1/2 when x = π/3 and x = 5π/3, and cos(x) = -1 when x = π, so our solutions are x = π/3, x = 5π/3 and x = π

Answer 2

Factor this. GCF=2

2(2cos²x+cos x-1)=0

2(2cos x -1)(cos x+1)=0 Set each factor equal to zero and solve

2cos x -1=0 Add 1 to both sides

2cos x =1 Divide by 2

cos x =1/2

x=π/3

x=5π/3

cos x+1=0 Subtract 1 from both sides

cos x=-1

x=π