Solve 2cos^2θ/2 = cos2θ for 0º≤θ≤360º. (Hint: Graph each side of the equation and find the points of intersection.)
A) ≈73.8º; 260.4º
B) ≈141.3º; 218.7º
C) ≈225.7º; 315.8º
D) ≈51.7º; 138.2º
+130458 Using cos^(x/2) = [1 + cos(x)] / 2
Then
2 * [1 + cos(x)] / 2 = cos(2x) →
1 + cos(x) = cos(2x) →
And using cos(2x) = 2cos^2(x) - 1
1 + cosx = 2cos^2(x) - 1 →
2cos^2(x) - cosx - 2 = 0
let x = cosx
2x^2 -x - 2 = 0 .....using the on-site solver......
2×x2−x−2=0⇒{x=−(√17−1)4x=(√17+1)4}⇒{x=−0.7807764064044151x=1.2807764064044151}
So, using the cosine inverse, and ignoring the second answer, we have.....
cos-1 [(1 - √17)/4] = x ...... so x = 141.331° and x = 218.668°
+130458 Using cos^(x/2) = [1 + cos(x)] / 2
Then
2 * [1 + cos(x)] / 2 = cos(2x) →
1 + cos(x) = cos(2x) →
And using cos(2x) = 2cos^2(x) - 1
1 + cosx = 2cos^2(x) - 1 →
2cos^2(x) - cosx - 2 = 0
let x = cosx
2x^2 -x - 2 = 0 .....using the on-site solver......
2×x2−x−2=0⇒{x=−(√17−1)4x=(√17+1)4}⇒{x=−0.7807764064044151x=1.2807764064044151}
So, using the cosine inverse, and ignoring the second answer, we have.....
cos-1 [(1 - √17)/4] = x ...... so x = 141.331° and x = 218.668°
