Help me!!! Algebra 2

Question

solve each equation for the given domain:

1)4cos²θ=1 for 0°≤θ≤360°

2)cos²θ+cosθ-1=0 for 0°≤θ≤180°

Answer 1

1) 4 cos^2 x = 1

cos^2x x= 1/4

cos x= + 1/2. and -1/2

cos x equals 1/2 at 60 and 300 degrees and -1/2 at 120 and 240 degrees

2) 2cos^2(x) + cos(x) -1 = 0 (factor this like any other quadratic )

(2cosx-1)(cosx+1)= 0. so we have 2 solutions

2cosx-1=0 cosx = -1 x =180degrees

2cosx=1

cos x =1/2

so x=60 degrees

Answer 2

We can solve these equations with a simple substitution:

For both problems, let x = cos θ. These two problems now become

1) 4x² = 1

2) x² + x - 1 = 0

Now we are back in algebra territory. The first equation 4x² = 1 has solution x = ± ¹/₂. (I trust you can fill in the details.) Since x = cos θ, we have cos θ = ± ¹/₂ . For 0°≤θ≤360°, we have four values of θ on the unit circle with cos θ = ± ¹/₂. The solution set is therefore {60°, 120°, 240°, 300°}.

For the second problem, let's solve x² + x - 1 = 0 by completing the square:

x² + x - 1 = 0 <----> x² + x = 1

<----> x² + x + (¹/₂)² = 1 + (¹/₂)²

<----> (x + ¹/₂)² = ⁵/₄

<----> x = ^{(-1 ± √5)}/₂

Remember our substitution x = cos θ ? This means cos θ = ^{(-1 ± √5)}/₂. Unfortunately, this doesn't match any coordinate listed on the unit circle. So, our final answer is the solution set {cos^{-1 (-1 + √5)}/₂, cos^{-1 (-1 - √5)}/₂}. For a numerical solution, you must plug these values into a calculator.

Answer 3

Cos² = 1/4. So cos =± 1/2. Angles are 60 120 240 and 300

for b)

use quadratic formula. a=1 b=1 c=-1

cosθ = -1±√1+4 / 2 = .618 or -1.6

cos has to between -1 and 1 always so .618 only works so cos^-1(.618) = 51.8